Known operations with logarithms. Natural logarithm, ln x function

basic properties.

logax + logay = log(x y);
logax − logay = log(x: y).

same grounds

log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

What if there is a degree in the base or argument of the logarithm? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x >

Task. Find the value of the expression:

Transition to a new foundation

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

Task. Find the value of the expression:

Basic properties of the logarithm

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Tolstoy.

Basic properties of logarithms

Knowing this rule you will know and exact value exhibitors, and the date of birth of Leo Tolstoy.

Examples for logarithms

Take the logarithm of expressions

Example 1
a). x=10ac^2 (a>0, c>0).

By properties 3,5 we calculate

4. where .

Example 2 Find x if

Example 3. Let the value of logarithms be given

Calculate log(x) if

Basic properties of logarithms

Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.

These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

Consider two logarithms with the same base: logax and logay. Then they can be added and subtracted, and:

logax + logay = log(x y);
logax − logay = log(x: y).

So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note: the key point here is - same grounds. If the bases are different, these rules do not work!

These formulas will help calculate the logarithmic expression even when its individual parts are not considered (see the lesson "What is a logarithm"). Take a look at the examples and see:

Since the bases of logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

Task. Find the value of the expression: log2 48 − log2 3.

The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 − log3 5.

Again, the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.

As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Based on this fact, many test papers. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

It is easy to see that last rule follows the first two. But it's better to remember it anyway - in some cases it will significantly reduce the amount of calculations.

Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers before the sign of the logarithm into the logarithm itself. This is what is most often required.

Task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument according to the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

Task. Find the value of the expression:

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:

I think to last example clarification is required. Where have logarithms gone? Until the very last moment, we work only with the denominator.

Formulas of logarithms. Logarithms are examples of solutions.

They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator have the same number: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result is the answer: 2.

Transition to a new foundation

Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the bases are different? What if they are not exact powers of the same number?

Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

In particular, if we put c = x, we get:

It follows from the second formula that it is possible to interchange the base and the argument of the logarithm, but in this case the whole expression is “turned over”, i.e. the logarithm is in the denominator.

These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when deciding logarithmic equations and inequalities.

However, there are tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:

Task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let's flip the second logarithm:

Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.

Task. Find the value of the expression: log9 100 lg 3.

The base and argument of the first logarithm are exact powers. Let's write it down and get rid of the indicators:

Now let's get rid of decimal logarithm, moving to a new base:

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:

In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called like this:

Indeed, what will happen if the number b is raised to such a degree that the number b in this degree gives the number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.

Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the value of the expression:

Note that log25 64 = log5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:

If someone is not in the know, this was a real task from the Unified State Examination 🙂

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.

logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument is one - the logarithm zero! Because a0 = 1 is a direct consequence of the definition.

That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out and solve the problems.

Basic properties of the logarithm

The above properties need to be known, since, on their basis, almost all problems and examples are solved based on logarithms. The remaining exotic properties can be derived by mathematical manipulations with these formulas

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

When calculating the formulas for the sum and difference of logarithms (3.4) are encountered quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.

Common cases of logarithms

Some of the common logarithms are those in which the base is even ten, exponential or deuce.
The base ten logarithm is usually called the base ten logarithm and is simply denoted lg(x).

It can be seen from the record that the basics are not written in the record. For example

natural logarithm is the logarithm of which is based on the exponent (denoted ln(x)).

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

And another important base two logarithm is

The derivative of the logarithm of the function is equal to one divided by the variable

The integral or antiderivative logarithm is determined by the dependence

The above material is enough for you to solve a wide class of problems related to logarithms and logarithms. For the sake of understanding the material, I will give only a few common examples from school curriculum and universities.

Examples for logarithms

Take the logarithm of expressions

Example 1
a). x=10ac^2 (a>0, c>0).

By properties 3,5 we calculate

2.
By the difference property of logarithms, we have

3.
Using properties 3.5 we find

4. where .

A seemingly complex expression using a series of rules is simplified to the form

Finding Logarithm Values

Example 2 Find x if

Decision. For the calculation, we apply properties 5 and 13 up to the last term

Substitute in the record and mourn

Since the bases are equal, we equate the expressions

Logarithms. First level.

Let the value of the logarithms be given

Calculate log(x) if

Solution: Take the logarithm of the variable to write the logarithm through the sum of the terms

This is just the beginning of acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the acquired knowledge to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another no less important topic- logarithmic inequalities ...

Basic properties of logarithms

These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

Consider two logarithms with the same base: logax and logay. Then they can be added and subtracted, and:

logax + logay = log(x y);
logax − logay = log(x: y).

These formulas will help calculate the logarithmic expression even when its individual parts are not considered (see the lesson "What is a logarithm"). Take a look at the examples and see:

Task. Find the value of the expression: log6 4 + log6 9.

Since the bases of logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

Task. Find the value of the expression: log2 48 − log2 3.

The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 − log3 5.

Again, the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.

As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

It is easy to see that the last rule follows their first two. But it's better to remember it anyway - in some cases it will significantly reduce the amount of calculations.

How to solve logarithms

This is what is most often required.

Task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument according to the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

Task. Find the value of the expression:

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:

I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator. They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.

Transition to a new foundation

Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:

Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

In particular, if we put c = x, we get:

These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.

However, there are tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:

Task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let's flip the second logarithm:

Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.

Task. Find the value of the expression: log9 100 lg 3.

The base and argument of the first logarithm are exact powers. Let's write it down and get rid of the indicators:

Now let's get rid of the decimal logarithm by moving to a new base:

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:

In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called like this:

Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the value of the expression:

Note that log25 64 = log5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:

If someone is not in the know, this was a real task from the Unified State Examination 🙂

Logarithmic unit and logarithmic zero

logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 1 is a direct consequence of the definition.

That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out and solve the problems.

The listed equalities when converting expressions with logarithms are used both from right to left and from left to right.

It is worth noting that it is not necessary to memorize the consequences of the properties: when carrying out transformations, you can get by with the basic properties of logarithms and other facts (for example, those for b≥0), from which the corresponding consequences follow. The "side effect" of this approach is only that the solution will be a little longer. For example, in order to do without the consequence, which is expressed by the formula , and starting only from the basic properties of logarithms, you will have to carry out a chain of transformations of the following form: .

The same can be said about the last property from the above list, which corresponds to the formula , since it also follows from the basic properties of logarithms. The main thing to understand is that it is always possible for the degree of a positive number with a logarithm in the exponent to swap the base of the degree and the number under the sign of the logarithm. In fairness, we note that examples involving the implementation of transformations of this kind are rare in practice. We will give a few examples below.

Converting numeric expressions with logarithms

We remembered the properties of logarithms, now it's time to learn how to put them into practice to transform expressions. It is natural to start with the transformation of numeric expressions, and not expressions with variables, since it is more convenient and easier to learn the basics on them. So we will do, and we will start with a very simple examples to learn how to choose the desired property of the logarithm, but we will gradually complicate the examples, up to the moment when several properties will need to be applied in a row to obtain the final result.

Selecting the desired property of logarithms

There are not so few properties of logarithms, and it is clear that you need to be able to choose the appropriate one from them, which in this particular case will lead to the desired result. Usually this is not difficult to do by comparing the form of the logarithm or expression being converted with the types of the left and right parts of the formulas expressing the properties of logarithms. If left or right part one of the formulas coincides with the given logarithm or expression, then, most likely, it is this property that should be used during the transformation. The following examples clearly demonstrate this.

Let's start with examples of transforming expressions using the definition of the logarithm, which corresponds to the formula a log a b =b , a>0 , a≠1 , b>0 .

Example.

Calculate, if possible: a) 5 log 5 4 , b) 10 log(1+2 π) , c) , d) 2 log 2 (−7) , e) .

Decision.

In the example, letter a) clearly shows the structure a log a b , where a=5 , b=4 . These numbers satisfy the conditions a>0 , a≠1 , b>0 , so you can safely use the equality a log a b =b . We have 5 log 5 4=4 .

b) Here a=10 , b=1+2 π , conditions a>0 , a≠1 , b>0 are fulfilled. In this case, the equality 10 lg(1+2 π) =1+2 π takes place.

c) And in this example we are dealing with a degree of the form a log a b , where and b=ln15 . So .

Despite belonging to the same form a log a b (here a=2 , b=−7 ), the expression under the letter d) cannot be converted by the formula a log a b =b . The reason is that it doesn't make sense because it contains a negative number under the logarithm sign. Moreover, the number b=−7 does not satisfy the condition b>0 , which makes it impossible to resort to the formula a log a b =b , since it requires the conditions a>0 , a≠1 , b>0 . So, we can't talk about computing the value 2 log 2 (−7) . In this case, writing 2 log 2 (−7) = −7 would be an error.

Similarly, in the example under the letter e) it is impossible to give a solution of the form , since the original expression does not make sense.

Answer:

a) 5 log 5 4 =4 , b) 10 log(1+2 π) =1+2 π , c) , d), e) expressions do not make sense.

It is often useful to transform positive number is represented as a power of some positive and non-one number with a logarithm in the exponent. It is based on the same definition of the logarithm a log a b =b , a>0 , a≠1 , b>0 , but the formula is applied from right to left, that is, in the form b=a log a b . For example, 3=e ln3 or 5=5 log 5 5 .

Let's move on to using the properties of logarithms to transform expressions.

Example.

Find the value of the expression: a) log −2 1, b) log 1 1, c) log 0 1, d) log 7 1, e) ln1, f) lg1, g) log 3.75 1, h) log 5 π 7 1 .

Decision.

In the examples under letters a), b) and c), the expressions log −2 1 , log 1 1 , log 0 1 are given, which do not make sense, since the base of the logarithm should not contain a negative number, zero or one, because we have defined logarithm only for a positive and non-unit base. Therefore, in examples a) - c) there can be no question of finding the value of the expression.

In all other tasks, obviously, the bases of the logarithms contain positive and non-unit numbers 7 , e , 10 , 3.75 and 5 π 7 respectively, and units are everywhere under the signs of the logarithms. And we know the property of the logarithm of unity: log a 1=0 for any a>0 , a≠1 . Thus, the values of expressions b) - f) are equal to zero.

Answer:

a), b), c) the expressions do not make sense, d) log 7 1=0, e) ln1=0, f) log1=0, g) log 3.75 1=0, h) log 5 e 7 1=0 .

Example.

Calculate: a) , b) lne , c) lg10 , d) log 5 π 3 −2 (5 π 3 −2), e) log −3 (−3) , f) log 1 1 .

Decision.

It is clear that we have to use the property of the logarithm of the base, which corresponds to the formula log a a=1 for a>0 , a≠1 . Indeed, in tasks under all letters, the number under the sign of the logarithm coincides with its base. Thus, I want to say right away that the value of each of the given expressions is 1 . However, do not rush to conclusions: in tasks under the letters a) - d) the values of the expressions are really equal to one, and in tasks e) and f) the original expressions do not make sense, so it cannot be said that the values of these expressions are equal to 1.

Answer:

a) , b) lne=1 , c) lg10=1 , d) log 5 π 3 −2 (5 π 3 −2)=1, e), f) expressions do not make sense.

Example.

Find the value: a) log 3 3 11 , b) , c) , d) log −10 (−10) 6 .

Decision.

Obviously, under the signs of the logarithms are some degrees of base. Based on this, we understand that the property of the degree of the base is useful here: log a a p =p, where a>0, a≠1 and p is any real number. Considering this, we have the following results: a) log 3 3 11 =11 , b) , in) . Is it possible to write a similar equality for the example under the letter d) of the form log −10 (−10) 6 =6? No, you can't, because log −10 (−10) 6 doesn't make sense.

Answer:

a) log 3 3 11 =11, b) , in) d) the expression does not make sense.

Example.

Express the expression as the sum or difference of logarithms in the same base: a) , b) , c) log((−5) (−12)) .

Decision.

a) The product is under the sign of the logarithm, and we know the property of the logarithm of the product log a (x y)=log a x+log a y , a>0 , a≠1 , x>0 , y>0 . In our case, the number in the base of the logarithm and the numbers in the product are positive, that is, they satisfy the conditions of the selected property, therefore, we can safely apply it: .

b) Here we use the property of the logarithm of the quotient , where a>0 , a≠1 , x>0 , y>0 . In our case, the base of the logarithm is a positive number e, the numerator and denominator π are positive, which means they satisfy the conditions of the property, so we have the right to use the chosen formula: .

c) First, note that the expression lg((−5) (−12)) makes sense. But at the same time, we do not have the right to apply the formula for the logarithm of the product log a (x y)=log a x+log a y , a>0 , a≠1 , x>0 , y>0 , since the numbers −5 and −12 are negative and do not satisfy the conditions x>0 , y>0 . That is, it is impossible to carry out such a transformation: log((−5)(−12))=log(−5)+log(−12). But what to do? In such cases, the original expression needs to be pre-transformed to avoid negative numbers. Pro similar cases expression conversions with negative numbers under the sign of the logarithm, we will talk in detail in one of, but for now we will give a solution to this example, which is clear in advance and without explanation: lg((−5)(−12))=lg(5 12)=lg5+lg12.

Answer:

a) , b) , c) lg((−5) (−12))=lg5+lg12 .

Example.

Simplify the expression: a) log 3 0.25 + log 3 16 + log 3 0.5, b) .

Decision.

Here, all the same properties of the logarithm of the product and the logarithm of the quotient that we used in the previous examples will help us, only now we will apply them from right to left. That is, we convert the sum of logarithms to the logarithm of the product, and the difference of the logarithms to the logarithm of the quotient. We have
a) log 3 0.25+log 3 16+log 3 0.5=log 3 (0.25 16 0.5)=log 3 2.
b) .

Answer:

a) log 3 0.25+log 3 16+log 3 0.5=log 3 2, b) .

Example.

Get rid of the degree under the sign of the logarithm: a) log 0.7 5 11, b) , c) log 3 (−5) 6 .

Decision.

It is easy to see that we are dealing with expressions like log a b p . The corresponding property of the logarithm is log a b p =p log a b , where a>0 , a≠1 , b>0 , p is any real number. That is, under the conditions a>0 , a≠1 , b>0 from the logarithm of the degree log a b p we can go to the product p·log a b . Let's carry out this transformation with the given expressions.

a) In this case a=0.7 , b=5 and p=11 . So log 0.7 5 11 =11 log 0.7 5 .

b) Here , the conditions a>0 , a≠1 , b>0 are fulfilled. So

c) The expression log 3 (−5) 6 has the same structure log a b p , a=3 , b=−5 , p=6 . But for b, the condition b>0 is not satisfied, which makes it impossible to apply the formula log a b p =p log a b . So why can't you get the job done? It is possible, but a preliminary transformation of the expression is required, which we will discuss in detail below in the paragraph under the heading . The solution will be like this: log 3 (−5) 6 =log 3 5 6 =6 log 3 5.

Answer:

a) log 0.7 5 11 =11 log 0.7 5 ,
b)
c) log 3 (−5) 6 =6 log 3 5 .

Quite often, the formula for the logarithm of the degree when carrying out transformations has to be applied from right to left in the form p log a b \u003d log a b p (this requires the same conditions for a, b and p). For example, 3 ln5=ln5 3 and lg2 log 2 3=log 2 3 lg2 .

Example.

a) Calculate the value of log 2 5 if it is known that lg2≈0.3010 and lg5≈0.6990. b) Write the fraction as a logarithm to base 3.

Decision.

a) The formula for the transition to a new base of the logarithm allows us to represent this logarithm as a ratio of decimal logarithms, the values of which are known to us: . It remains only to carry out the calculations, we have .

b) Here it is enough to use the formula for the transition to a new base, and apply it from right to left, that is, in the form . We get .

Answer:

a) log 2 5≈2.3223, b) .

At this stage, we have rather scrupulously considered the transformation of the simplest expressions using the basic properties of logarithms and the definition of a logarithm. In these examples, we had to use one property and nothing else. Now, with a clear conscience, you can move on to examples whose transformation requires the use of several properties of logarithms and other additional transformations. We will deal with them in the next paragraph. But before that, let us briefly dwell on examples of the application of consequences from the basic properties of logarithms.

Example.

a) Get rid of the root under the sign of the logarithm. b) Convert the fraction to a base 5 logarithm. c) Get rid of the powers under the sign of the logarithm and at its base. d) Calculate the value of the expression . e) Replace the expression with a power with base 3.

Decision.

a) If we recall the corollary from the property of the logarithm of the degree , then you can immediately answer: .

b) Here we use the formula from right to left, we have .

c) In this case, the formula leads to the result . We get .

d) And here it suffices to apply the corollary to which the formula corresponds . So .

e) The property of the logarithm allows us to achieve the desired result: .

Answer:

a) . b) . in) . G) . e) .

Consistently Applying Multiple Properties

Real tasks for transforming expressions using the properties of logarithms are usually more complicated than those that we dealt with in the previous paragraph. In them, as a rule, the result is not obtained in one step, but the solution already consists in the sequential application of one property after another, together with additional identical transformations, such as opening brackets, reducing like terms, reducing fractions, etc. So let's get closer to such examples. There is nothing complicated about this, the main thing is to act carefully and consistently, observing the order in which the actions are performed.

Example.

Calculate the value of an expression (log 3 15−log 3 5) 7 log 7 5.

Decision.

The difference of logarithms in brackets by the property of the logarithm of the quotient can be replaced by the logarithm log 3 (15:5) , and then calculate its value log 3 (15:5)=log 3 3=1 . And the value of the expression 7 log 7 5 by the definition of the logarithm is 5 . Substituting these results into the original expression, we get (log 3 15−log 3 5) 7 log 7 5 =1 5=5.

Here is a solution without explanation:
(log 3 15−log 3 5) 7 log 7 5 =log 3 (15:5) 5=
= log 3 3 5=1 5=5 .

Answer:

(log 3 15−log 3 5) 7 log 7 5 =5.

Example.

What is the value of the numerical expression log 3 log 2 2 3 −1 ?

Decision.

Let's first transform the logarithm, which is under the sign of the logarithm, according to the formula of the logarithm of the degree: log 2 2 3 =3. So log 3 log 2 2 3 =log 3 3 and then log 3 3=1 . So log 3 log 2 2 3 −1=1−1=0 .

Answer:

log 3 log 2 2 3 −1=0 .

Example.

Simplify the expression.

Decision.

The formula for converting to a new base of the logarithm allows the ratio of logarithms to one base to be represented as log 3 5 . In this case, the original expression will take the form . By definition of the logarithm 3 log 3 5 =5 , that is , and the value of the resulting expression, by virtue of the same definition of the logarithm, is equal to two.

Here short version solution, which is usually given: .

Answer:

For a smooth transition to the information of the next paragraph, let's take a look at the expressions 5 2+log 5 3 , and lg0.01 . Their structure does not fit any of the properties of logarithms. So what happens if they cannot be converted using the properties of logarithms? It is possible if you carry out preliminary transformations that prepare these expressions for applying the properties of logarithms. So 5 2+log 5 3 =5 2 5 log 5 3 =25 3=75, and lg0,01=lg10 −2 = −2 . Further we will understand in detail how such preparation of expressions is carried out.

Preparing expressions to apply the properties of logarithms

Logarithms in the converted expression very often differ in the structure of the notation from the left and right parts of formulas that correspond to the properties of logarithms. But just as often, the transformation of these expressions involves the use of the properties of logarithms: their use only requires preliminary preparation. And this preparation consists in carrying out certain identical transformations that bring logarithms to a form convenient for applying properties.

In fairness, we note that almost any transformation of expressions can act as preliminary transformations, from the banal reduction of similar terms to the application trigonometric formulas. This is understandable, since the converted expressions can contain any mathematical objects: brackets, modules, fractions, roots, degrees, etc. Thus, one must be prepared to perform any required transformation in order to further benefit from the properties of logarithms.

Let's say right away that in this section we do not set ourselves the task of classifying and analyzing all conceivable preliminary transformations that allow us to apply the properties of logarithms or the definition of a logarithm in the future. Here we will focus on only four of them, which are the most characteristic and most often encountered in practice.

And now in detail about each of them, after which, within the framework of our topic, it remains only to deal with the transformation of expressions with variables under the signs of logarithms.

Selection of powers under the sign of the logarithm and in its base

Let's start right away with an example. Let us have a logarithm. Obviously, in this form, its structure is not conducive to the use of the properties of logarithms. Is it possible to somehow transform this expression in order to simplify it, or even better calculate its value? To answer this question, let's take a closer look at the numbers 81 and 1/9 in the context of our example. It is easy to see here that these numbers can be represented as a power of 3 , indeed, 81=3 4 and 1/9=3 −2 . In this case, the original logarithm is presented in the form and it becomes possible to apply the formula . So, .

The analysis of the analyzed example gives rise to the following idea: if possible, you can try to highlight the degree under the sign of the logarithm and at its base in order to apply the property of the logarithm of the degree or its consequence. It remains only to figure out how to single out these degrees. We will give some recommendations on this issue.

Sometimes it is quite obvious that the number under the sign of the logarithm and / or in its base represents some integer power, as in the example discussed above. Almost constantly you have to deal with powers of two, which are well familiar: 4=2 2 , 8=2 3 , 16=2 4 , 32=2 5 , 64=2 6 , 128=2 7 , 256=2 8 , 512= 2 9 , 1024=2 10 . The same can be said about the degrees of the triple: 9=3 2 , 27=3 3 , 81=3 4 , 243=3 5 , ... In general, it does not hurt if there is degree table natural numbers within ten. It is also not difficult to work with integer powers of ten, hundred, thousand, etc.

Example.

Calculate the value or simplify the expression: a) log 6 216 , b) , c) log 0.000001 0.001 .

Decision.

a) Obviously, 216=6 3 , so log 6 216=log 6 6 3 =3 .

b) The table of powers of natural numbers allows us to represent the numbers 343 and 1/243 as powers of 7 3 and 3 −4, respectively. Therefore, the following transformation of the given logarithm is possible:

c) Since 0.000001=10 −6 and 0.001=10 −3, then log 0.000001 0.001=log 10 −6 10 −3 =(−3)/(−6)=1/2.

Answer:

a) log 6 216=3, b) , c) log 0.000001 0.001=1/2 .

In more difficult cases to highlight the powers of numbers, you have to resort to.

Example.

Convert expression to more plain sight log 3 648 log 2 3 .

Decision.

Let's see what the expansion of the number 648 into prime factors:

That is, 648=2 3 3 4 . Thus, log 3 648 log 2 3=log 3 (2 3 3 4) log 2 3.

Now we convert the logarithm of the product to the sum of logarithms, after which we apply the properties of the logarithm of the degree:
log 3 (2 3 3 4) log 2 3=(log 3 2 3 + log 3 3 4) log 2 3=
=(3 log 3 2+4) log 2 3 .

By virtue of the corollary of the property of the logarithm of the degree, which corresponds to the formula , the product log32 log23 is the product , and it is known to be equal to one. Considering this, we get 3 log 3 2 log 2 3+4 log 2 3=3 1+4 log 2 3=3+4 log 2 3.

Answer:

log 3 648 log 2 3=3+4 log 2 3.

Quite often, expressions under the sign of the logarithm and in its base are products or ratios of the roots and / or powers of some numbers, for example, , . Similar expressions can be represented as a degree. To do this, the transition from roots to degrees is carried out, and and are applied. These transformations allow you to select the degrees under the sign of the logarithm and in its base, and then apply the properties of logarithms.

Example.

Calculate: a) , b).

Decision.

a) The expression in the base of the logarithm is the product of powers with the same bases, by the corresponding property of powers we have 5 2 5 −0.5 5 −1 =5 2−0.5−1 =5 0.5.

Now let's convert the fraction under the sign of the logarithm: let's move from the root to the degree, after which we will use the property of the ratio of degrees with the same bases: .

It remains to substitute the results obtained into the original expression, use the formula and finish the transformation:

b) Since 729=3 6 , and 1/9=3 −2 , the original expression can be rewritten as .

Next, apply the property of the root of the exponent, move from the root to the exponent, and use the ratio property of the powers to convert the base of the logarithm to a power: .

Taking into account the last result, we have .

Answer:

a) , b).

It is clear that in the general case, to obtain powers under the sign of the logarithm and in its base, various transformations of various expressions may be required. Let's give a couple of examples.

Example.

What is the value of the expression: a) , b) .

Decision.

Further, we note that the given expression has the form log A B p , where A=2 , B=x+1 and p=4 . We transformed numerical expressions of this kind according to the property of the logarithm of the degree log a b p \u003d p log a b, therefore, with a given expression, I want to do the same, and from log 2 (x + 1) 4 go to 4 log 2 (x + 1) . And now let's calculate the value of the original expression and the expression obtained after the transformation, for example, with x=−2 . We have log 2 (−2+1) 4 =log 2 1=0 , and 4 log 2 (−2+1)=4 log 2 (−1)- meaningless expression. This raises a legitimate question: “What did we do wrong”?

And the reason is as follows: we performed the transformation log 2 (x+1) 4 =4 log 2 (x+1) , based on the formula log a b p =p log a b , but we have the right to apply this formula only if the conditions a >0 , a≠1 , b>0 , p - any real number. That is, the transformation we have done takes place if x+1>0 , which is the same x>−1 (for A and p, the conditions are met). However, in our case, the ODZ of the variable x for the original expression consists not only of the interval x> −1 , but also of the interval x<−1 . Но для x<−1 мы не имели права осуществлять преобразование по выбранной формуле.

The need to take into account ODZ

Let's continue to analyze the transformation of the expression log 2 (x+1) 4 we have chosen, and now let's see what happens to the ODZ when passing to the expression 4 log 2 (x+1) . In the previous paragraph, we found the ODZ of the original expression - this is the set (−∞, −1)∪(−1, +∞) . Now let's find the area of acceptable values of the variable x for the expression 4 log 2 (x+1) . It is determined by the condition x+1>0 , which corresponds to the set (−1, +∞) . It is obvious that when going from log 2 (x+1) 4 to 4·log 2 (x+1), the range of admissible values narrows. And we agreed to avoid reforms that lead to a narrowing of the ODZ, as this can lead to various negative consequences.

Here it is worth noting for yourself that it is useful to control the ODZ at each step of the transformation and not allow it to narrow. And if suddenly at some stage of the transformation there was a narrowing of the ODZ, then it is worth looking very carefully at whether this transformation is permissible and whether we had the right to carry it out.

In fairness, we say that in practice we usually have to work with expressions in which the ODZ of variables is such that it allows us to use the properties of logarithms without restrictions in the form already known to us, both from left to right and from right to left, when carrying out transformations. You quickly get used to this, and you begin to carry out the transformations mechanically, without thinking about whether it was possible to carry them out. And at such moments, as luck would have it, more complex examples slip through, in which the inaccurate application of the properties of logarithms leads to errors. So you need to always be on the alert, and make sure that there is no narrowing of the ODZ.

It does not hurt to separately highlight the main transformations based on the properties of logarithms, which must be carried out very carefully, which can lead to a narrowing of the DPV, and as a result, to errors:

Some transformations of expressions according to the properties of logarithms can also lead to the opposite - the expansion of the ODZ. For example, going from 4 log 2 (x+1) to log 2 (x+1) 4 extends the ODZ from the set (−1, +∞) to (−∞, −1)∪(−1, +∞) . Such transformations take place if you remain within the ODZ for the original expression. So the transformation just mentioned 4 log 2 (x+1)=log 2 (x+1) 4 takes place on the ODZ variable x for the original expression 4 log 2 (x+1) , that is, when x+1> 0 , which is the same as (−1, +∞) .

Now that we have discussed the nuances that you need to pay attention to when converting expressions with variables using the properties of logarithms, it remains to figure out how these conversions should be carried out correctly.

X+2>0 . Does it work in our case? To answer this question, let's take a look at the DPV of the x variable. It is determined by the system of inequalities , which is equivalent to the condition x+2>0 (if necessary, see the article solution of systems of inequalities). Thus, we can safely apply the property of the logarithm of the degree.

We have
3 log(x+2) 7 −log(x+2)−5 log(x+2) 4 =
=3 7 log(x+2)−log(x+2)−5 4 log(x+2)=
=21 log(x+2)−log(x+2)−20 log(x+2)=
=(21−1−20)lg(x+2)=0 .

You can act differently, since the ODZ allows you to do this, for example like this:

Answer:

3 log(x+2) 7 −log(x+2)−5 log(x+2) 4 =0.

And what to do when the conditions associated with the properties of logarithms are not met on the ODZ? We will deal with this with examples.

Let us be required to simplify the expression lg(x+2) 4 −lg(x+2) 2 . The transformation of this expression, unlike the expression from the previous example, does not allow the free use of the property of the logarithm of the degree. Why? The ODZ of the variable x in this case is the union of two intervals x>−2 and x<−2 . При x>−2 we can safely apply the property of the logarithm of the degree and proceed as in the example above: log(x+2) 4 −log(x+2) 2 =4 log(x+2)−2 log(x+2)=2 log(x+2). But the ODZ contains another interval x+2<0 , для которого последнее преобразование будет некорректно. Что же делать при x+2<0 ? В подобных случаях на помощь приходит . Определение модуля позволяет выражение x+2 при x+2<0 представить как −|x+2| . Тогда при x+2<0 от lg(x+2) 4 −lg(x+2) 2 переходим к log(−|x+2|) 4 −log(−|x+2|) 2 and further, due to the power properties of lg|x+2| 4−lg|x+2| 2. The resulting expression can be transformed according to the property of the logarithm of the degree, since |x+2|>0 for any values of the variable. We have log|x+2| 4−lg|x+2| 2 =4 log|x+2|−2 log|x+2|=2 log|x+2|. Now you can get rid of the module, since it has done its job. Since we are transforming at x+2<0 , то 2·lg|x+2|=2·lg(−(x+2)) . Итак, можно считать, что мы справились с поставленной задачей. Ответ: . Полученный результат можно записать компактно с использованием модуля как .

Let's consider one more example to make working with modules familiar. Let us conceive from the expression pass to the sum and difference of the logarithms of the linear binomials x−1 , x−2 and x−3 . First we find the ODZ:

On the interval (3, +∞), the values of the expressions x−1 , x−2 and x−3 are positive, so we can safely apply the properties of the logarithm of the sum and difference:

And on the interval (1, 2), the values of the expression x−1 are positive, and the values of the expressions x−2 and x−3 are negative. Therefore, on the interval under consideration, we represent x−2 and x−3 using the modulo as −|x−2| and −|x−3| respectively. Wherein

Now we can apply the properties of the logarithm of the product and the quotient, since on the considered interval (1, 2) the values of the expressions x−1 , |x−2| and |x−3| - positive.

We have

The results obtained can be combined:

In general, similar reasoning allows, based on the formulas for the logarithm of the product, ratio and degree, to obtain three practically useful results that are quite convenient to use:

The logarithm of the product of two arbitrary expressions X and Y of the form log a (X·Y) can be replaced by the sum of logarithms log a |X|+log a |Y| , a>0 , a≠1 .
The special logarithm log a (X:Y) can be replaced by the difference of the logarithms log a |X|−log a |Y| , a>0 , a≠1 , X and Y are arbitrary expressions.
From the logarithm of some expression B to an even power p of the form log a B p, one can pass to the expression p log a |B| , where a>0 , a≠1 , p is an even number and B is an arbitrary expression.

Similar results are given, for example, in instructions for solving exponential and logarithmic equations in the collection of problems in mathematics for applicants to universities, edited by M. I. Skanavi.

Example.

Simplify the expression .

Decision.

It would be good to apply the properties of the logarithm of the degree, sum and difference. But can we do it here? To answer this question, we need to know the ODZ.

Let's define it:

It is quite obvious that the expressions x+4 , x−2 and (x+4) 13 on the range of possible values of the variable x can take both positive and negative values. Therefore, we will have to work through modules.

Module properties allow you to rewrite as , so

Also, nothing prevents you from using the property of the logarithm of the degree, and then bring like terms:

Another sequence of transformations leads to the same result:

and since the expression x−2 can take both positive and negative values on the ODZ, when taking an even exponent 14

These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

Consider two logarithms with the same base: log a x and log a y. Then they can be added and subtracted, and:

log a x+log a y= log a (x · y);
log a x−log a y= log a (x : y).

These formulas will help you calculate the logarithmic expression even when its individual parts are not considered (see the lesson "What is a logarithm"). Take a look at the examples and see:

log 6 4 + log 6 9.

Since the bases of logarithms are the same, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.

Task. Find the value of the expression: log 2 48 − log 2 3.

The bases are the same, we use the difference formula:
log 2 48 - log 2 3 = log 2 (48: 3) = log 2 16 = 4.

Task. Find the value of the expression: log 3 135 − log 3 5.

Again, the bases are the same, so we have:
log 3 135 − log 3 5 = log 3 (135: 5) = log 3 27 = 3.

Removing the exponent from the logarithm

It is easy to see that the last rule follows their first two. But it's better to remember it anyway - in some cases it will significantly reduce the amount of calculations.

Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x> 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers before the sign of the logarithm into the logarithm itself. This is what is most often required.

Task. Find the value of the expression: log 7 49 6 .

Let's get rid of the degree in the argument according to the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the value of the expression:
[Figure caption]

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 2 4 ; 49 = 72. We have:

[Figure caption]

Now let's look at the main fraction. The numerator and denominator have the same number: log 2 7. Since log 2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result is the answer: 2.

Transition to a new foundation

Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:

Let the logarithm log a x. Then for any number c such that c> 0 and c≠ 1, the equality is true:
[Figure caption]
In particular, if we put c = x, we get:
[Figure caption]

These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.

However, there are tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:

Task. Find the value of the expression: log 5 16 log 2 25.

Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;

Now let's flip the second logarithm:

[Figure caption]

Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.

Task. Find the value of the expression: log 9 100 lg 3.

The base and argument of the first logarithm are exact powers. Let's write it down and get rid of the indicators:

[Figure caption]

Now let's get rid of the decimal logarithm by moving to a new base:

[Figure caption]

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:

In the first case, the number n becomes the exponent of the argument. Number n can be absolutely anything, because it's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called the basic logarithmic identity.

Indeed, what will happen if the number b raise to the power so that b to this extent gives a number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.

Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the value of the expression:
[Figure caption]

Note that log 25 64 = log 5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:

[Figure caption]

If someone is not in the know, this was a real task from the exam :)

Logarithmic unit and logarithmic zero

log a a= 1 is the logarithmic unit. Remember once and for all: the logarithm to any base a from this base itself is equal to one.
log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument is one, the logarithm is zero! because a 0 = 1 is a direct consequence of the definition.

That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out and solve the problems.

Problem B7 gives an expression that needs to be simplified. The result should be common number which can be written on the answer sheet. All expressions are conditionally divided into three types:

logarithmic,
Demonstration,
Combined.

Demonstration and logarithmic expressions in pure form practically do not occur. However, knowing how they are calculated is essential.

In general, problem B7 is solved quite simply and is quite within the power of the average graduate. The lack of clear algorithms is compensated by its standard and uniformity. You can learn how to solve such problems simply by a large number workouts.

Logarithmic Expressions

The vast majority of B7 problems contain logarithms in one form or another. This topic is traditionally considered difficult, since it is usually studied in the 11th grade - the era of mass preparation for final exams. As a result, many graduates have a very vague idea about logarithms.

But in this task, no one requires deep theoretical knowledge. We will meet only the simplest expressions that require straightforward reasoning and may well be mastered independently. Below are the basic formulas you need to know to deal with logarithms:

In addition, one must be able to replace roots and fractions with powers with a rational exponent, otherwise in some expressions there will simply be nothing to take out from under the sign of the logarithm. Replacement formulas:

Task. Find expression values:
log 6 270 − log 6 7.5
log 5 775 − log 5 6.2

The first two expressions are converted as the difference of logarithms:
log 6 270 − log 6 7.5 = log 6 (270: 7.5) = log 6 36 = 2;
log 5 775 − log 5 6.2 = log 5 (775: 6.2) = log 5 125 = 3.

To calculate the third expression, you will have to select degrees - both in the base and in the argument. First, let's find the internal logarithm:

Then - external:

Constructions like log a log b x seem complicated and misunderstood to many. Meanwhile, this is just the logarithm of the logarithm, i.e. log a (log b x ). First, the inner logarithm is calculated (put log b x = c ), and then the outer one: log a c .

exponential expressions

We will call an exponential expression any construction of the form a k , where the numbers a and k are arbitrary constants, and a > 0. Methods for working with such expressions are quite simple and are considered in the 8th grade algebra lessons.

Below are the basic formulas that you must know. The application of these formulas in practice, as a rule, does not cause problems.

a n a m = a n + m ;
a n / a m = a n − m ;
(a n ) m = a n m ;
(a b) n = a n b n ;
(a : b ) n = a n : b n .

If a complex expression with powers is encountered, and it is not clear how to approach it, a universal technique is used - decomposition into prime factors. As a result big numbers in the bases of degrees are replaced by simple and understandable elements. Then it remains only to apply the above formulas - and the problem will be solved.

Task. Find expression values: 7 9 3 11: 21 8 , 24 7: 3 6: 16 5 , 30 6: 6 5: 25 2 .

Decision. We decompose all bases of powers into prime factors:
7 9 3 11: 21 8 = 7 9 3 11: (7 3) 8 = 7 9 3 11: (7 8 3 8) = 7 9 3 11: 7 8: 3 8 = 7 3 3 = 189.
24 7: 3 6: 16 5 = (3 2 3) 7: 3 6: (2 4) 5 = 3 7 2 21: 3 6: 2 20 = 3 2 = 6.
30 6: 6 5: 25 2 = (5 3 2) 6: (3 2) 5: (5 2) 2 = 5 6 3 6 2 6: 3 5: 2 5: 5 4 = 5 2 3 2 = 150.

Combined tasks

If you know the formulas, then all exponential and logarithmic expressions are solved literally in one line. However, in problem B7, powers and logarithms can be combined to form rather strong combinations.

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