The absolute number of the logarithm. What is a logarithm

\(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)

Let's explain it easier. For example, \(\log_(2)(8)\) is equal to the power \(2\) must be raised to to get \(8\). From this it is clear that \(\log_(2)(8)=3\).

Examples:		\(\log_(5)(25)=2\)		because \(5^(2)=25\)
		\(\log_(3)(81)=4\)		because \(3^(4)=81\)
		\(\log_(2)\)\(\frac(1)(32)\) \(=-5\)		because \(2^(-5)=\)\(\frac(1)(32)\)

Argument and base of the logarithm

Any logarithm has the following "anatomy":

The argument of the logarithm is usually written at its level, and the base is written in subscript closer to the sign of the logarithm. And this entry is read like this: "the logarithm of twenty-five to the base of five."

How to calculate the logarithm?

To calculate the logarithm, you need to answer the question: to what degree should the base be raised to get the argument?

For example, calculate the logarithm: a) \(\log_(4)(16)\) b) \(\log_(3)\)\(\frac(1)(3)\) c) \(\log_(\sqrt (5))(1)\) d) \(\log_(\sqrt(7))(\sqrt(7))\) e) \(\log_(3)(\sqrt(3))\)

a) To what power must \(4\) be raised to get \(16\)? Obviously the second. That's why:

\(\log_(4)(16)=2\)

\(\log_(3)\)\(\frac(1)(3)\) \(=-1\)

c) To what power must \(\sqrt(5)\) be raised to get \(1\)? And what degree makes any number a unit? Zero, of course!

\(\log_(\sqrt(5))(1)=0\)

d) To what power must \(\sqrt(7)\) be raised to get \(\sqrt(7)\)? In the first - any number in the first degree is equal to itself.

\(\log_(\sqrt(7))(\sqrt(7))=1\)

e) To what power must \(3\) be raised to get \(\sqrt(3)\)? From we know what is fractional degree, which means Square root is the degree \(\frac(1)(2)\) .

\(\log_(3)(\sqrt(3))=\)\(\frac(1)(2)\)

Example : Calculate the logarithm \(\log_(4\sqrt(2))(8)\)

Solution :

\(\log_(4\sqrt(2))(8)=x\)		We need to find the value of the logarithm, let's denote it as x. Now let's use the definition of the logarithm: \(\log_(a)(c)=b\) \(\Leftrightarrow\) \(a^(b)=c\)
\((4\sqrt(2))^(x)=8\)		What links \(4\sqrt(2)\) and \(8\)? Two, because both numbers can be represented by twos: \(4=2^(2)\) \(\sqrt(2)=2^(\frac(1)(2))\) \(8=2^(3)\)
\(((2^(2)\cdot2^(\frac(1)(2))))^(x)=2^(3)\)		On the left, we use the degree properties: \(a^(m)\cdot a^(n)=a^(m+n)\) and \((a^(m))^(n)=a^(m\cdot n)\)
\(2^(\frac(5)(2)x)=2^(3)\)		The bases are equal, we proceed to the equality of indicators
\(\frac(5x)(2)\) \(=3\)		Multiply both sides of the equation by \(\frac(2)(5)\)
		The resulting root is the value of the logarithm

Answer : \(\log_(4\sqrt(2))(8)=1,2\)

Why was the logarithm invented?

To understand this, let's solve the equation: \(3^(x)=9\). Just match \(x\) to make the equality work. Of course, \(x=2\).

Now solve the equation: \(3^(x)=8\). What is x equal to? That's the point.

The most ingenious will say: "X is a little less than two." How exactly is this number to be written? To answer this question, they came up with the logarithm. Thanks to him, the answer here can be written as \(x=\log_(3)(8)\).

I want to emphasize that \(\log_(3)(8)\), as well as any logarithm is just a number. Yes, it looks unusual, but it is short. Because if we wanted to write it in the form decimal fraction, then it would look like this: \(1.892789260714.....\)

Example : Solve the equation \(4^(5x-4)=10\)

Solution :

\(4^(5x-4)=10\)		\(4^(5x-4)\) and \(10\) cannot be reduced to the same base. So here you can not do without the logarithm. Let's use the definition of the logarithm: \(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)
\(\log_(4)(10)=5x-4\)		Flip the equation so x is on the left
\(5x-4=\log_(4)(10)\)		Before us. Move \(4\) to the right. And don't be afraid of the logarithm, treat it like a regular number.
\(5x=\log_(4)(10)+4\)		Divide the equation by 5
\(x=\)\(\frac(\log_(4)(10)+4)(5)\)		Here is our root. Yes, it looks unusual, but the answer is not chosen.

Answer : \(\frac(\log_(4)(10)+4)(5)\)

Decimal and natural logarithms

As stated in the definition of the logarithm, its base can be any positive number except one \((a>0, a\neq1)\). And among all possible grounds there are two that occur so often that a special short notation was invented for logarithms with them:

Natural logarithm: a logarithm whose base is the Euler number \(e\) (equal to approximately \(2.7182818…\)), and the logarithm is written as \(\ln(a)\).

I.e, \(\ln(a)\) is the same as \(\log_(e)(a)\)

Decimal logarithm: A logarithm whose base is 10 is written \(\lg(a)\).

I.e, \(\lg(a)\) is the same as \(\log_(10)(a)\), where \(a\) is some number.

Basic logarithmic identity

Logarithms have many properties. One of them is called "Basic logarithmic identity" and looks like this:

\(a^(\log_(a)(c))=c\)

This property follows directly from the definition. Let's see how this formula came about.

Let's remember short note logarithm definitions:

if \(a^(b)=c\), then \(\log_(a)(c)=b\)

That is, \(b\) is the same as \(\log_(a)(c)\). Then we can write \(\log_(a)(c)\) instead of \(b\) in the formula \(a^(b)=c\) . It turned out \(a^(\log_(a)(c))=c\) - the main logarithmic identity.

You can find the rest of the properties of logarithms. With their help, you can simplify and calculate the values ​​of expressions with logarithms, which are difficult to calculate directly.

Example : Find the value of the expression \(36^(\log_(6)(5))\)

Solution :

Answer : \(25\)

How to write a number as a logarithm?

As mentioned above, any logarithm is just a number. The converse is also true: any number can be written as a logarithm. For example, we know that \(\log_(2)(4)\) is equal to two. Then you can write \(\log_(2)(4)\) instead of two.

But \(\log_(3)(9)\) is also equal to \(2\), so you can also write \(2=\log_(3)(9)\) . Similarly with \(\log_(5)(25)\), and with \(\log_(9)(81)\), etc. That is, it turns out

\(2=\log_(2)(4)=\log_(3)(9)=\log_(4)(16)=\log_(5)(25)=\log_(6)(36)=\ log_(7)(49)...\)

Thus, if we need, we can write the two as a logarithm with any base anywhere (even in an equation, even in an expression, even in an inequality) - we just write the squared base as an argument.

It's the same with a triple - it can be written as \(\log_(2)(8)\), or as \(\log_(3)(27)\), or as \(\log_(4)(64) \) ... Here we write the base in the cube as an argument:

\(3=\log_(2)(8)=\log_(3)(27)=\log_(4)(64)=\log_(5)(125)=\log_(6)(216)=\ log_(7)(343)...\)

And with four:

\(4=\log_(2)(16)=\log_(3)(81)=\log_(4)(256)=\log_(5)(625)=\log_(6)(1296)=\ log_(7)(2401)...\)

And with minus one:

\(-1=\) \(\log_(2)\)\(\frac(1)(2)\) \(=\) \(\log_(3)\)\(\frac(1)( 3)\) \(=\) \(\log_(4)\)\(\frac(1)(4)\) \(=\) \(\log_(5)\)\(\frac(1 )(5)\) \(=\) \(\log_(6)\)\(\frac(1)(6)\) \(=\) \(\log_(7)\)\(\frac (1)(7)\)\(...\)

And with one third:

\(\frac(1)(3)\) \(=\log_(2)(\sqrt(2))=\log_(3)(\sqrt(3))=\log_(4)(\sqrt( 4))=\log_(5)(\sqrt(5))=\log_(6)(\sqrt(6))=\log_(7)(\sqrt(7))...\)

Any number \(a\) can be represented as a logarithm with base \(b\): \(a=\log_(b)(b^(a))\)

Example : Find the value of an expression \(\frac(\log_(2)(14))(1+\log_(2)(7))\)

Solution :

Answer : \(1\)

One of the elements of primitive level algebra is the logarithm. The name came from Greek from the word “number” or “power” and means the power to which it is necessary to raise the number at the base to find the final number.

Types of logarithms

• log a b is the logarithm of the number b to the base a (a > 0, a ≠ 1, b > 0);
• lg b - decimal logarithm (logarithm base 10, a = 10);
• ln b - natural logarithm (logarithm base e, a = e).

How to solve logarithms?

The logarithm of the number b to the base a is an exponent, which requires that the base a be raised to the number b. The result is pronounced like this: “logarithm of b to the base of a”. The solution to logarithmic problems is that you need to determine the given degree by the numbers by the specified numbers. There are some basic rules for determining or solving the logarithm, as well as transforming the notation itself. Using them, a solution is made logarithmic equations, derivatives are found, integrals are solved, and many other operations are performed. Basically, the solution to the logarithm itself is its simplified notation. Below are the main formulas and properties:

For any a ; a > 0; a ≠ 1 and for any x ; y > 0.

• a log a b = b is the basic logarithmic identity
• log a 1 = 0
• log a a = 1
• log a (x y ) = log a x + log a y
• log a x/ y = log a x – log a y
• log a 1/x = -log a x
• log a x p = p log a x
• log a k x = 1/k log a x , for k ≠ 0
• log a x = log a c x c
• log a x \u003d log b x / log b a - formula for the transition to a new base
• log a x = 1/log x a

How to solve logarithms - step by step instructions for solving

• First, write down the required equation.

Please note: if the base logarithm is 10, then the record is shortened, a decimal logarithm is obtained. If worth natural number e, then we write down, reducing to a natural logarithm. It means that the result of all logarithms is the power to which the base number is raised to obtain the number b.

Directly, the solution lies in the calculation of this degree. Before solving an expression with a logarithm, it must be simplified according to the rule, that is, using formulas. You can find the main identities by going back a little in the article.

Adding and subtracting logarithms with two various numbers, but with the same bases, replace with one logarithm with the product or division of the numbers b and c, respectively. In this case, you can apply the transition formula to another base (see above).

If you are using expressions to simplify the logarithm, there are some limitations to be aware of. And that is: the base of the logarithm a is only a positive number, but not equal to one. The number b, like a, must be greater than zero.

There are cases when, having simplified the expression, you will not be able to calculate the logarithm in numerical form. It happens that such an expression does not make sense, because many degrees are irrational numbers. Under this condition, leave the power of the number as a logarithm.

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: log a x and log a y. Then they can be added and subtracted, and:

1. log a x+log a y= log a (x · y);
2. log a x−log a y= log a (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. Note: key moment here - same grounds. If the bases are different, these rules do not work!

These formulas will help you calculate 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.

A 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.

A 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.

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

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.

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.

A 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

A 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]

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.

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

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 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]

It follows from the second formula that the base and the argument of the logarithm can be interchanged, but 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 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:

A 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.

A 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.

A 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

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.

1. 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.
2. 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.

Today we will talk about logarithm formulas and give demonstration solution examples.

By themselves, they imply solution patterns according to the basic properties of logarithms. Before applying the logarithm formulas to the solution, we recall for you, first all the properties:

Now, based on these formulas (properties), we show examples of solving logarithms.

Examples of solving logarithms based on formulas.

Logarithm a positive number b in base a (denoted log a b) is the exponent to which a must be raised to get b, with b > 0, a > 0, and 1.

According to the definition log a b = x, which is equivalent to a x = b, so log a a x = x.

Logarithms, examples:

log 2 8 = 3, because 2 3 = 8

log 7 49 = 2 because 7 2 = 49

log 5 1/5 = -1, because 5 -1 = 1/5

Decimal logarithm is an ordinary logarithm, the base of which is 10. Denoted as lg.

log 10 100 = 2 because 10 2 = 100

natural logarithm- also the usual logarithm logarithm, but already with the base e (e \u003d 2.71828 ... - irrational number). Referred to as ln.

It is desirable to remember the formulas or properties of logarithms, because we will need them later when solving logarithms, logarithmic equations and inequalities. Let's work through each formula again with examples.

• Basic logarithmic identity
  a log a b = b

  8 2log 8 3 = (8 2log 8 3) 2 = 3 2 = 9

• The logarithm of the product is equal to the sum of the logarithms
  log a (bc) = log a b + log a c

  log 3 8.1 + log 3 10 = log 3 (8.1*10) = log 3 81 = 4

• The logarithm of the quotient is equal to the difference of the logarithms
  log a (b/c) = log a b - log a c

  9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81

• Properties of the degree of a logarithmable number and the base of the logarithm

  The exponent of a logarithm number log a b m = mlog a b

  Exponent of the base of the logarithm log a n b =1/n*log a b

  log a n b m = m/n*log a b,

  if m = n, we get log a n b n = log a b

  log 4 9 = log 2 2 3 2 = log 2 3

• Transition to a new foundation
  log a b = log c b / log c a,

  if c = b, we get log b b = 1

  then log a b = 1/log b a

  log 0.8 3*log 3 1.25 = log 0.8 3*log 0.8 1.25/log 0.8 3 = log 0.8 1.25 = log 4/5 5/4 = -1

As you can see, the logarithm formulas are not as complicated as they seem. Now, having considered examples of solving logarithms, we can move on to logarithmic equations. We will consider examples of solving logarithmic equations in more detail in the article: "". Do not miss!

If you still have questions about the solution, write them in the comments to the article.

Note: decided to get an education of another class study abroad as an option.

In relation to

the task of finding any of the three numbers from the other two, given, can be set. Given a and then N is found by exponentiation. If N are given and then a is found by extracting the root of the power x (or exponentiation). Now consider the case when, given a and N, it is required to find x.

Let the number N be positive: the number a is positive and not equal to one: .

Definition. The logarithm of the number N to the base a is the exponent to which you need to raise a to get the number N; the logarithm is denoted by

Thus, in equality (26.1), the exponent is found as the logarithm of N to the base a. Entries

have the same meaning. Equality (26.1) is sometimes called the basic identity of the theory of logarithms; in fact, it expresses the definition of the concept of the logarithm. By this definition the base of the logarithm a is always positive and different from unity; the logarithmable number N is positive. Negative numbers and zero do not have logarithms. It can be proved that any number with a given base has a well-defined logarithm. Therefore equality entails . Note that the condition is essential here, otherwise the conclusion would not be justified, since the equality is true for any values ​​of x and y.

Example 1. Find

Solution. To get the number, you need to raise base 2 to the power Therefore.

You can record when solving such examples in the following form:

Example 2. Find .

Solution. We have

In examples 1 and 2, we easily found the desired logarithm by representing the logarithmable number as a degree of base with a rational exponent. IN general case, for example, for etc., this cannot be done, since the logarithm has an irrational value. Let us pay attention to one question related to this statement. In § 12 we gave the concept of the possibility of determining any real power of a given positive number. This was necessary for the introduction of logarithms, which, in general, can be irrational numbers.

Consider some properties of logarithms.

Property 1. If the number and base are equal, then the logarithm is equal to one, and, conversely, if the logarithm is equal to one, then the number and base are equal.

Proof. Let By the definition of the logarithm, we have and whence

Conversely, let Then by definition

Property 2. The logarithm of unity to any base is equal to zero.

Proof. By the definition of the logarithm (the zero power of any positive base is equal to one, see (10.1)). From here

Q.E.D.

The converse statement is also true: if , then N = 1. Indeed, we have .

Before formulating next property logarithms, we agree to say that two numbers a and b lie on the same side of a third number c if they are both either greater than c or less than c. If one of these numbers is greater than c and the other is less than c, then we will say that they lie on opposite sides of c.

Property 3. If the number and base lie on the same side of unity, then the logarithm is positive; if the number and base lie on opposite sides of unity, then the logarithm is negative.

The proof of property 3 is based on the fact that the degree of a is greater than one if the base is greater than one and the exponent is positive, or the base is less than one and the exponent is negative. The degree is less than one if the base is greater than one and the exponent is negative, or the base is less than one and the exponent is positive.

There are four cases to be considered:

We confine ourselves to the analysis of the first of them, the reader will consider the rest on his own.

Let then the exponent in equality be neither negative nor equal to zero, therefore, it is positive, i.e., which was required to be proved.

Example 3. Find out which of the following logarithms are positive and which are negative:

Solution, a) since the number 15 and the base 12 are located on the same side of the unit;

b) , since 1000 and 2 are located on the same side of the unit; at the same time, it is not essential that the base is greater than the logarithmic number;

c), since 3.1 and 0.8 lie on opposite sides of unity;

G) ; why?

e) ; why?

The following properties 4-6 are often called the rules of logarithm: they allow, knowing the logarithms of some numbers, to find the logarithms of their product, quotient, degree of each of them.

Property 4 (the rule for the logarithm of the product). The logarithm of the product of several positive numbers in a given base is equal to the sum of the logarithms of these numbers in the same base.

Proof. Let positive numbers be given.

For the logarithm of their product, we write the equality (26.1) defining the logarithm:

From here we find

Comparing the exponents of the first and last expressions, we obtain the required equality:

Note that the condition is essential; logarithm of the product of two negative numbers makes sense, but in this case we get

In general, if the product of several factors is positive, then its logarithm is equal to the sum of the logarithms of the modules of these factors.

Property 5 (quotient logarithm rule). The logarithm of a quotient of positive numbers is equal to the difference between the logarithms of the dividend and the divisor, taken in the same base. Proof. Consistently find

Q.E.D.

Property 6 (rule of the logarithm of the degree). The logarithm of the power of any positive number is equal to the logarithm of that number times the exponent.

Proof. We write again the main identity (26.1) for the number :

Q.E.D.

Consequence. The logarithm of the root of a positive number is equal to the logarithm of the root number divided by the exponent of the root:

We can prove the validity of this corollary by presenting how and using property 6.

Example 4. Logarithm to base a:

a) (it is assumed that all values ​​b, c, d, e are positive);

b) (it is assumed that ).

Solution, a) It is convenient to pass in this expression to fractional powers:

Based on equalities (26.5)-(26.7) we can now write:

We notice that simpler operations are performed on the logarithms of numbers than on the numbers themselves: when multiplying numbers, their logarithms are added, when divided, they are subtracted, etc.

That is why logarithms have been used in computational practice (see Sec. 29).

The action inverse to the logarithm is called potentiation, namely: potentiation is the action by which this number itself is found by the given logarithm of a number. In essence, potentiation is not any special action: it comes down to raising the base to a power ( equal to the logarithm numbers). The term "potentiation" can be considered synonymous with the term "exponentiation".

When potentiating, it is necessary to use the rules that are inverse to the rules of logarithm: replace the sum of logarithms with the logarithm of the product, the difference of logarithms with the logarithm of the quotient, etc. In particular, if there is any factor in front of the sign of the logarithm, then during potentiation it must be transferred to the indicator degrees under the sign of the logarithm.

Example 5. Find N if it is known that

Solution. In connection with the potentiation rule just stated, the factors 2/3 and 1/3, which are in front of the signs of the logarithms on the right side of this equality, will be transferred to the exponents under the signs of these logarithms; we get

Now we replace the difference of logarithms with the logarithm of the quotient:

to obtain the last fraction in this chain of equalities, we freed the previous fraction from irrationality in the denominator (section 25).

Property 7. If the base is greater than one, then more has a larger logarithm (and a smaller one has a smaller one), if the base is less than one, then a larger number has a smaller logarithm (and a smaller one has a larger one).

This property is also formulated as a rule for the logarithm of inequalities, both parts of which are positive:

When taking the logarithm of inequalities with a base greater than one, the inequality sign is preserved, and when taking a logarithm with a base less than one, the sign of the inequality is reversed (see also item 80).

The proof is based on properties 5 and 3. Consider the case when If , then and, taking the logarithm, we obtain

(a and N/M lie on the same side of unity). From here

Case a follows, the reader will figure it out for himself.