Logarithms with different indicators. What is logarithm

The B7 task is given a certain expression that you need to simplify. As a result, the usual number should be written in the replies form. All expressions are conventionally divided into three types:

1. Logarithmic
2. Indicative
3. Combined.

Indicative and logarithmic expressions in pure form are practically not found. However, it is absolutely necessary to know how they are calculated.

In general, the task B7 is solved quite simply and quite under the middle graduate. The lack of clear algorithms is compensated for in it standards and monotiance. Learning to solve such tasks can be simply due to a large number of training.

Logarithmic expressions

The overwhelming majority of B7 tasks contain logarithms in one form or another. This topic is traditionally considered difficult, since its study is necessary, as a rule, on the 11th grade - the era of mass preparation for final exams. As a result, many graduates have a very vague idea of \u200b\u200blogarithms.

But in this problem no one does not require deep theoretical knowledge. We will meet only the simplest expressions that require uncomplicated reasoning and may well be mastered on their own. Below are the basic formulas that you need to know to cope with logarithms:

In addition, it is necessary to be able to replace the roots and fractions to the degree with a rational indicator, otherwise in some expressions, it is simply nothing to carry out the logarithm. Replacement formulas:

A task. Find expressions:
LOG 6 270 - LOG 6 7.5
log 5 775 - LOG 5 6.2

The first two expressions are converted as a difference in logarithms:
Log 6 270 - log 6 7.5 \u003d log 6 (270: 7.5) \u003d log 6 36 \u003d 2;
Log 5 775 - log 5 6.2 \u003d log 5 (775: 6,2) \u003d log 5 125 \u003d 3.

To calculate the third expression, it will be necessary to allocate degrees - both at the base and in the argument. To begin with, we will find the internal logarithm:

Then - external:

The design of the Log A log b x seems complex and incomprehensible to many. Meanwhile, it is just a logarithm from logarithm, i.e. Log A (log b x). First, the internal logarithm is calculated (put Log B x \u003d c), and then external: Log A C.

Indicative expressions

Let us be called an indicative expression any design of the form a k, where the numbers a and k are arbitrary constants, and a\u003e 0. Methods of working with such expressions are quite simple and considered in the lessons of the 8th class algebra.

Below are the basic formulas that must be known. The use of these formulas in practice, as a rule, does not cause problems.

1. a n · a m \u003d a n + m;
2. a n / a m \u003d a n - m;
3. (a n) m \u003d a n · m;
4. (a · b) n \u003d a n · b n;
5. (A: B) n \u003d a n: b n.

If a complex expression was met with degrees, and it is not clear how to approach it, a universal reception is used - decomposition on simple factors. As a result, large numbers in the bases of degrees are replaced by simple and understandable elements. Then it will only be used to apply the above formulas - and the task will be solved.

A task. Find values \u200b\u200bof expressions: 7 9 · 3 11: 21 8, 24 7: 3 6: 16 5, 30 6: 6 5: 25 2.

Decision. Spread all the foundations of degrees to ordinary multipliers:
7 9 · 3 11: 21 8 \u003d 7 9 · 3 11: (7 · 3) 8 \u003d 7 9 · 3 11: (7 8 · 3 8) \u003d 7 9 · 3 11: 7 8: 3 8 \u003d 7 · 3 3 \u003d 189.
24 7: 3 6: 16 5 \u003d (3 · 2 3) 7: 3 6: (2 4) 5 \u003d 3 7 · 2 21: 3 6: 2 20 \u003d 3 · 2 \u003d 6.
30 6: 6 5: 25 2 \u003d (5 · 3 · 2) 6: (3 · 2) 5: (5 2) 2 \u003d 5 6 · 3 6 · 2 6: 3 5: 2 5: 5 4 \u003d 5 2 · 3 · 2 \u003d 150.

Combined tasks

If you know the formula, then all the indicative and logarithmic expressions are solved literally in one line. However, in the problem B7, the degree and logarithms can be combined, forming rather unreasonable combinations.

Tasks whose solution lies in transformation of logarithmic expressions, quite often meet on the exam.

To successfully cope with them at minimal time except the main logarithmic identities, you need to know and correctly use some formulas.

This is: a log a b \u003d b, where a, b\u003e 0, and ≠ 1 (it follows directly from the definition of the logarithm).

log a b \u003d log with b / log with a or log a b \u003d 1 / log b a
where a, b, c\u003e 0; A, C ≠ 1.

log a m b n \u003d (m / n) log | a | | b |
where a, b\u003e 0, a ≠ 1, m, n є r, n ≠ 0.

and log with b \u003d b log with a
where a, b, c\u003e 0 and a, b, s ≠ 1

To show the fairness of the fourth equality, prologate the left and the right-hand side of the A. We get Log A (and log with b) \u003d log A (b log with a) or log with B \u003d log with a · log a b; log with b \u003d log with a · (log with b / log with a); Log with B \u003d log with b.

We have proven equality of logarithms, which means equal to the expressions under logarithms. Formula 4 is proved.

Example 1.

Calculate 81 LOG 27 5 LOG 5 4.

Decision.

81 = 3 4 , 27 = 3 3 .

log 27 5 \u003d 1/3 log 3 5, log 5 4 \u003d log 3 4 / LOG 3 5. Consequently,

log 27 5 · log 5 4 \u003d 1/3 log 3 5 · (Log 3 4 / Log 3 5) \u003d 1/3 log 3 4.

Then 81 log 27 5 Log 5 4 \u003d (3 4) 1/3 log 3 4 \u003d (3 log 3 4) 4/3 \u003d (4) 4/3 \u003d 4 3 √4.

You can independently perform the following task.

Calculate (8 log 2 3 + 3 1 / Log 2 3) - log 0.2 5.

As a tip 0.2 \u003d 1/5 \u003d 5 -1; Log 0.2 5 \u003d -1.

Answer: 5.

Example 2.

Calculate (√11) log. √3 9- log 121 81.

Decision.

Perform the replacement of expressions: 9 \u003d 3 2, √3 \u003d 3 1/2, Log √3 9 \u003d 4,

121 \u003d 11 2, 81 \u003d 3 4, Log 121 81 \u003d 2 log 11 3 (Formula 3 was used).

Then (√11) Log √3 9- log 121 81 \u003d (11 1/2) 4-2 log 11 3 \u003d (11) 2- log 11 3 \u003d 11 2 / (11) Log 11 3 \u003d 11 2 / ( 11 LOG 11 3) \u003d 121/3.

Example 3.

Calculate Log 2 24 / Log 96 2- log 2 192 / log 12 2.

Decision.

Logarithms contained in the example, replace logarithms with a base 2.

log 96 2 \u003d 1 / log 2 96 \u003d 1 / log 2 (2 5 · 3) \u003d 1 / (Log 2 2 5 + log 2 3) \u003d 1 / (5 + log 2 3);

log 2 192 \u003d log 2 (2 6 · 3) \u003d (log 2 2 6 + log 2 3) \u003d (6 + log 2 3);

log 2 24 \u003d log 2 (2 3 · 3) \u003d (log 2 2 3 + log 2 3) \u003d (3 + LOG 2 3);

log 12 2 \u003d 1 / log 2 12 \u003d 1 / log 2 (2 2 · 3) \u003d 1 / (Log 2 2 2 + log 2 3) \u003d 1 / (2 + log 2 3).

Then log 2 24 / log 96 2 - log 2 192 / log 12 2 \u003d (3 + log 2 3) / (1 / (5 + log 2 3)) - ((6 + LOG 2 3) / (1 / ( 2 + log 2 3)) \u003d

\u003d (3 + Log 2 3) · (5 + log 2 3) - (6 + log 2 3) (2 + Log 2 3).

After disclosing brackets and bringing similar terms, we obtain the number 3. (with a simplification of the expression, Log 2 3 can designate via n and simplify expression

(3 + n) · (5 + n) - (6 + n) (2 + n)).

Answer: 3.

You can independently perform the following task:

Calculate (log 3 4 + log 4 3 + 2) · Log 3 16 · log 2 144 3.

Here it is necessary to make a transition to logarithms based on 3 and decomposition on simple multipliers of large numbers.

Answer: 1/2.

Example 4.

Three numbers a \u003d 1 / (log 3 0.5), B \u003d 1 / (log 0.5 3), C \u003d log 0.5 12 - log 0.5 3. Place them in ascending order.

Decision.

We transform the numbers a \u003d 1 / (log 3 0.5) \u003d log 0.5 3; C \u003d LOG 0.5 12 - log 0.5 3 \u003d log 0.5 12/3 \u003d log 0.5 4 \u003d -2.

Compare them

log 0.5 3\u003e log 0.5 4 \u003d -2 and log 0.5 3< -1 = log 0,5 2, так как функция у = log 0,5 х – убывающая.

Or 2< log 0,5 3 < -1. Тогда -1 < 1/(log 0,5 3) < -1/2.

Answer. Consequently, the procedure for placing numbers: C; BUT; IN.

Example 5.

How many integers are located on the interval (Log 3 1/16; log 2 6 48).

Decision.

We define between what degrees of the number 3 is the number 1/16. We get 1/27< 1 / 16 < 1 / 9 .

Since the function y \u003d log 3 x is increasing, then log 3 (1/2 27)< log 3 (1 / 16) < log 3 (1 / 9); -3 < log 3 (1 / 16) < -2.

log 6 48 \u003d log 6 (36 · 4/3) \u003d log 6 36 + log 6 (4/3) \u003d 2 + log 6 (4/3). Compare Log 6 (4/3) and 1/5. And for this, compare numbers 4/3 and 6 1/5. Erected both numbers in 5 degree. We obtain (4/3) 5 \u003d 1024/243 \u003d 4 52/243< 6. Следовательно,

log 6 (4/3)< 1 / 5 . 2 < log 6 48 < 2 1 / 5 . Числа, входящие в двойное неравенство, положительные. Их можно возводить в квадрат. Знаки неравенства при этом не изменятся. Тогда 4 < log 6 2 48 < 4 21 / 25.

Consequently, the interval (log 3 1/16; log 6 48) includes the interval [-2; 4] and the integers are placed on it; -one; 0; one; 2; 3; four.

Answer: 7 integers.

Example 6.

Calculate 3 LGLG 2 / LG 3 - LG20.

Decision.

3 LG LG 2 / LG 3 \u003d (3 1 / LG3) LG LG 2 \u003d (3 Lo G 3 10) LG LG 2 \u003d 10 LG LG 2 \u003d LG2.

Then 3 lglg2 / lg3 - lg 20 \u003d lg 2 - lg 20 \u003d lg 0.1 \u003d -1.

Answer: -1.

Example 7.

It is known that Log 2 (√3 + 1) + log 2 (√6 - 2) \u003d A. Find Log 2 (√3 -1) + log 2 (√6 + 2).

Decision.

Numbers (√3 + 1) and (√3 - 1); (√6 - 2) and (√6 + 2) - conjugate.

We will conduct the following transformation of expressions

√3 - 1 \u003d (√3 - 1) · (√3 + 1)) / (√3 + 1) \u003d 2 / (√3 + 1);

√6 + 2 \u003d (√6 + 2) · (√6 - 2)) / (√6 - 2) \u003d 2 / (√6 - 2).

Then log 2 (√3 - 1) + Log 2 (√6 + 2) \u003d log 2 (2 / (√3 + 1)) + log 2 (2 / (√6 - 2)) \u003d

Log 2 2 - log 2 (√3 + 1) + log 2 2 - log 2 (√6 - 2) \u003d 1 - log 2 (√3 + 1) + 1 - Log 2 (√6 - 2) \u003d

2 - log 2 (√3 + 1) - Log 2 (√6 - 2) \u003d 2 - A.

Answer: 2 - A.

Example 8..

Simplify and find the approximate value of the expression (log 3 2 · log 4 3 · Log 5 4 · Log 6 5 · ... · Log 10 9.

Decision.

All logarithms we give to the total base 10.

(log 3 2 · log 4 3 · log 5 4 · log 6 5 · ... · log 10 9 \u003d (LG 2 / LG 3) · (LG 3 / LG 4) · (LG 4 / LG 5) · (LG 5 / LG 6) · ... · (LG 8 / LG 9) · LG 9 \u003d LG 2 ≈ 0.3010. (The approximate LG 2 value can be found using a table, a logarithmic line or a calculator).

Answer: 0,3010.

Example 9..

Calculate Log A 2 B 3 √ (A 11 B -3) if log √ A B 3 \u003d 1. (In this example, and 2 B 3 is the base of the logarithm).

Decision.

If log √ a b 3 \u003d 1, then 3 / (0.5 log a b \u003d 1. and log a B \u003d 1/6.

Then log a 2 B 3√ (A 11 b -3) \u003d 1/2 Log A 2 B 3 (A 11 b -3) \u003d Log A (A 11 B -3) / (2log A (A 2 B 3) ) \u003d (log A A 11 + Log A B -3) / (2 (log a A A 2 + LOG A B 3)) \u003d (11 - 3Log A B) / (2 (2 + 3Log A B)) Considering that that log a B \u003d 1/6 is obtained (11 - 3 · 1/6) / (2 (2 + 3 · 1/6)) \u003d 10.5 / 5 \u003d 2.1.

Answer: 2.1.

You can independently perform the following task:

Calculate Log √3 6 √2.1 if log 0.7 27 \u003d a.

Answer: (3 + a) / (3a).

Example 10.

Calculate 6.5 4 / LOG 3 169 · 3 1 / LOG 4 13 + LOG125.

Decision.

6.5 4 / Log 3 169 · 3 1 / log 4 13 + log 125 \u003d (13/2) 4/2 Log 3 13 · 3 2 / log 2 13 + 2Log 5 5 3 \u003d (13/2) 2 log 13 3 · 3 2 log 13 2 + 6 \u003d (13 log 13 3/2 Log 13 3) 2 · (3 log 13 2) 2 + 6 \u003d (3/2 log 13 3) 2 · (3 log 13 2) 2 + 6 \u003d (3 2 / (2 log 13 3) 2) · (2 \u200b\u200bLog 13 3) 2 + 6.

(2 Log 13 3 \u003d 3 log 13 2 (Formula 4))

We obtain 9 + 6 \u003d 15.

Answer: 15.

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We continue to learn logarithms. In this article we will talk about calculation of logarithm, this process is called logarithming. First we will deal with the calculation of logarithms by definition. Next, consider how the values \u200b\u200bof logarithms are located using their properties. After that, let's stop at the calculation of logarithms through the initially specified values \u200b\u200bof other logarithms. Finally, learn to use the tables of logarithms. All theory is equipped with examples with detailed solutions.

Navigating page.

Calculation of logarithms by definition

In the simplest cases, it is possible enough quickly and easily. finding logarithm by definition. Let's consider in detail how this process occurs.

Its essence consists in the representation of the number B in the form A C, from where to determine the logarithm number C is the value of the logarithm. That is, finding the logarithm by definition is answered by the following chain of equalities: Log A B \u003d Log A A C \u003d C.

So, the calculation of the logarithm by definition comes down to finding such a number C, which a C \u003d B, and the number C itself is the desired logarithm value.

Given the information of the previous paragraphs when the number under the sign of the logarithm is set by some degree of logarithm base, you can immediately indicate what is equal to the logarithm - it is equal to the degree. Let's show solutions for examples.

Example.

Find Log 2 2 -3, as well as calculate the natural logarithm of the number E 5.3.

Decision.

The definition of a logarithm allows us to immediately say that Log 2 2 -3 \u003d -3. Indeed, the number under the logarithm sign is equal to the base of 2 V -3 degree.

Similarly, we find the second logarithm: LNE 5.3 \u003d 5.3.

Answer:

log 2 2-3 \u003d -3 and LNE 5.3 \u003d 5.3.

If the number B under the logarithm sign is not specified as the degree of base of the logarithm, then it is necessary to carefully look if it is impossible to come to the representation of the number B in the form A C. Often such a representation is quite obvious, especially when the number under the logarithm sign is equal to the ground to the degree 1, or 2, or 3, ...

Example.

Calculate logarithms Log 5 25, and.

Decision.

It is easy to see that 25 \u003d 5 2, it allows you to calculate the first logarithm: log 5 25 \u003d log 5 5 2 \u003d 2.

Go to the calculation of the second logarithm. The number can be represented as a degree of number 7: (see if necessary). Hence, .

We rewrite the third logarithm in the following form. Now you can see that where we conclude that . Consequently, by definition of logarithm .

Briefly, the solution could be written like this :.

Answer:

log 5 25 \u003d 2, and .

When the logarithm sign is a rather large natural number, it will not hurt to decompose on simple multipliers. This often helps to present such a number as a certain degree of logarithm base, which means to calculate this logarithm by definition.

Example.

Find the logarithm value.

Decision.

Some logarithms allow you to immediately specify the value of logarithms. These properties include the property of the logarithm unit and the logarithm property of the number equal to the base: Log 1 1 \u003d Log A A 0 \u003d 0 and LOG A A \u003d Log A A 1 \u003d 1. That is, when the logarithm is the number 1 or number A, equal to the base of the logarithm, then in these cases, logarithms are 0 and 1, respectively.

Example.

What are the logarithms and LG10?

Decision.

Since, then from the definition of logarithm .

In the second example, the number 10 under the sign of the logarithm coincides with its base, so the decimal logarithm of ten is equal to one, that is, LG10 \u003d LG10 1 \u003d 1.

Answer:

AND lG10 \u003d 1.

Note that the calculation of logarithms by definition (which we disassembled in the previous paragraph) implies the use of the equality of Log A A P \u003d P, which is one of the properties of logarithms.

In practice, when the number under the logarithm sign and the base of logarithm is easily presented in the form of a degree of a certain number, it is very convenient to use the formula which corresponds to one of the properties of logarithms. Consider an example of finding a logarithm, illustrating the use of this formula.

Example.

Calculate logarithm.

Decision.

Answer:

.

Not mentioned above the properties of logarithms are also used when calculating, but we will talk about this in the following paragraphs.

Finding logarithms through other well-known logarithms

Information of this paragraph continues the use of the properties of logarithms when calculating them. But here the main difference is that the properties of logarithms are used in order to express the source logarithm through another logarithm, the value of which is known. Let us give an example for explanation. Suppose we know that log 2 3≈1.584963, then we can find, for example, log 2 6 by performing a small conversion using the logarithm properties: log 2 6 \u003d Log 2 (2 · 3) \u003d log 2 2 + log 2 3≈ 1+1,584963=2,584963 .

In the given example, we were enough to use the logarithm of the work. However, much more often it is necessary to apply a wider arsenal of the properties of logarithms to calculate the source logarithm through the specified.

Example.

Calculate the logarithm 27 for the base 60, if it is known that Log 60 2 \u003d A and LOG 60 5 \u003d b.

Decision.

So, we need to find Log 60 27. It is easy to see that 27 \u003d 3 3, and the source logarithm by virtue of the logarithm property of the degree can be rewritten as 3 · log 60 3.

Now let's see how log 60 3 express through well-known logarithms. The property of the logarithm of a number equal to the base allows you to record the equality LOG 60 60 \u003d 1. On the other hand, Log 60 60 \u003d Log60 (2 2 · 3 · 5) \u003d lOG 60 2 2 + LOG 60 3 + LOG 60 5 \u003d 2 · LOG 60 2 + LOG 60 3 + LOG 60 5. In this way, 2 · LOG 60 2 + LOG 60 3 + LOG 60 5 \u003d 1. Hence, log 60 3 \u003d 1-2 · log 60 2-log 60 5 \u003d 1-2 · A-B.

Finally, we calculate the source logarithm: log 60 27 \u003d 3 · log 60 3 \u003d 3 · (1-2 · a-b) \u003d 3-6 · a-3 · b.

Answer:

log 60 27 \u003d 3 · (1-2 · a-b) \u003d 3-6 · a-3 · b.

Separately, it is worth saying about the value of the formula for the transition to the new base of the logarithm . It allows from logarithms with any bases to move to logarithms with a specific basis, the values \u200b\u200bof which are known or have the ability to find them. Usually, the initial logarithm of the transition formula is transferred to logarithms on one of the bases 2, e or 10, since there are logarithms tables that allow them to calculate their values \u200b\u200bwith a certain degree of accuracy. In the next point, we will show how it is done.

Logarov tables, their use

To approximately calculate the values \u200b\u200bof logarithms can be used. tables Logarovmov. The most commonly used table of logarithms based on 2, the table of natural logarithms and a table of decimal logarithms. When working in a decimal number system, it is convenient to use the table of logarithms based on ten. With it, we will learn to find the values \u200b\u200bof logarithms.

The table presented allows with an accuracy of one ten thousand to find the values \u200b\u200bof decimal logarithms of numbers from 1,000 to 9.9999 (with three decimal plates). The principle of finding the logarithm value using a table of decimal logs will be analyzed on a specific example - so clearer. Find LG1,256.

In the left column of the decimal logarithms, we find the two first digits of the number 1.256, that is, we find 1.2 (this number is visited by the blue line). The third digit of the number 1.256 (digit 5) is found in the first or last row to the left of the double line (this number is circled with a red line). The fourth digit of the original number 1.256 (digit 6) is found in the first or last row to the right of the double line (this number is circled with a green line). Now we find the numbers in the logarithms tables on the intersection of the marked line and marked columns (these numbers are highlighted with orange). The sum of the marked numbers gives the desired value of the decimal logarithm with an accuracy of the fourth sign after the comma, that is, lG1,236≈0,0969 + 0.0021 \u003d 0,0990.

And is it possible, using the table below, find the values \u200b\u200bof decimal logarithms of numbers that have more than three digits after the comma, as well as leaving 1 to 9,999? Yes, you can. Let's show how this is done, on the example.

Calculate LG102,76332. First you need to write down standard: 102,76332 \u003d 1,0276332 · 10 2. After that, Mantissa should be rounded to the third mark after the comma, we have 1,0276332 · 10 2 ≈1,028 · 10 2At the same time, the initial decimal logarithm is approximately equal to the logarithm of the resulting number, that is, we accept LG102,76332≈LG1,76332 · 10 2. Now apply the logarithm properties: lG1,028 · 10 2 \u003d LG1,028 + LG10 2 \u003d LG1,028 + 2. Finally, we find the value of the logarithm LG1,028 on the table of decimal logarithms LG1,028≈0,0086 + 0.0034 \u003d 0.012. As a result, the entire process of calculating the logarithm looks like this: lG102,76332 \u003d LG1,0276332 · 10 2 ≈LG1,028 · 10 2 \u003d lG1,028 + LG10 2 \u003d LG1,028 + 2≈0,012 + 2 \u003d 2.012.

In conclusion, it is worth noting that using the table of decimal logs, you can calculate the approximate value of any logarithm. To do this, it is enough using the transition formula to go to decimal logarithms, find their values \u200b\u200bon the table, and perform the remaining calculations.

For example, calculate Log 2 3. By the transition formula to the new base of logarithm. From the table of decimal logs, we find LG3≈0,4771 and LG2≈0,30110. In this way, .

Bibliography.

• Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. et al. Algebra and start analysis: a textbook for 10 - 11 classes of general educational institutions.
• Gusev V.A., Mordkovich A.G. Mathematics (allowance for applicants to technical schools).

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What is logarithm?

Attention!
This topic has additional
Materials in a special section 555.
For those who are strongly "not very ..."
And for those who are "very ...")

What is logarithm? How to solve logarithms? These issues of many graduates are introduced into a stupor. Traditionally, the theme of logarithms is considered complex, incomprehensible and terrible. Especially - equations with logarithms.

It is absolutely wrong. Absolutely! Do not believe? Okay. Now for some 10 - 20 minutes you:

1. Catch what is logarithm.

2. Learn to solve a whole class of indicative equations. Even if nothing heard about them.

3. Learn to calculate simple logarithms.

And for this you will need to know only the multiplication table, but how the number is erected into a degree ...

I feel in doubt ... well, okay, set the time! Go!

To begin with, solve in mind here is such an equation:

If you like this site ...

By the way, I have another couple of interesting sites for you.)

It can be accessed in solving examples and find out your level. Testing with instant check. Learn - with interest!)

You can get acquainted with features and derivatives.