What is a logarithm for dummies. Logarithmic expressions. examples

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The logarithm of a positive number b to base a (a>0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a > 0, a ≠ 1, b > 0)       

Note that the logarithm of a non-positive number is undefined. In addition, the base of the logarithm must be positive number, not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the logarithm to the base -2 of 4 is equal to 2.

Basic logarithmic identity

a log a b = b (a > 0, a ≠ 1) (2)

It is important that the scope of definition of the right and left sides of this formula is different. Left side is defined only for b>0, a>0 and a ≠ 1. The right-hand side is defined for any b, and does not depend on a at all. Thus, the application of the basic logarithmic “identity” when solving equations and inequalities can lead to a change in the OD.

Two obvious consequences of the definition of logarithm

log a a = 1 (a > 0, a ≠ 1) (3)
log a 1 = 0 (a > 0, a ≠ 1) (4)

Indeed, when raising the number a to the first power, we get the same number, and when raising it to the first power zero degree- one.

Logarithm of the product and logarithm of the quotient

log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) (5)

Log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) (6)

I would like to warn schoolchildren against thoughtlessly using these formulas when solving logarithmic equations and inequalities. When using them “from left to right,” the ODZ narrows, and when moving from the sum or difference of logarithms to the logarithm of the product or quotient, the ODZ expands.

Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f(x) and g(x) are both less than zero.

Transforming this expression into the sum log a f (x) + log a g (x), we are forced to limit ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of acceptable values, and this is categorically unacceptable, since it can lead to a loss of solutions. A similar problem exists for formula (6).

The degree can be taken out of the sign of the logarithm

log a b p = p log a b (a > 0, a ≠ 1, b > 0) (7)

And again I would like to call for accuracy. Consider the following example:

Log a (f (x) 2 = 2 log a f (x)

The left side of the equality is obviously defined for all values ​​of f(x) except zero. The right side is only for f(x)>0! By taking the degree out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of acceptable values. All these remarks apply not only to power 2, but also to any even power.

Formula for moving to a new foundation

log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) (8)

That rare case, when the ODZ does not change during the transformation. If you have chosen base c wisely (positive and not equal to 1), the formula for moving to a new base is completely safe.

If we choose the number b as the new base c, we get an important special case formulas (8):

Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)

Some simple examples with logarithms

Example 1. Calculate: log2 + log50.
Solution. log2 + log50 = log100 = 2. We used the sum of logarithms formula (5) and the definition of the decimal logarithm.

Example 2. Calculate: lg125/lg5.
Solution. log125/log5 = log 5 125 = 3. We used the formula for moving to a new base (8).

Table of formulas related to logarithms

a log a b = b (a > 0, a ≠ 1)

log a a = 1 (a > 0, a ≠ 1)

log a 1 = 0 (a > 0, a ≠ 1)

log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0)

log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0)

log a b p = p log a b (a > 0, a ≠ 1, b > 0)

log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1)

log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1)

So, we have powers of two. If you take the number from the bottom line, you can easily find the power to which you will have to raise two to get this number. For example, to get 16, you need to raise two to the fourth power. And to get 64, you need to raise two to the sixth power. This can be seen from the table.

And now - actually, the definition of the logarithm:

The base a logarithm of x is the power to which a must be raised to get x.

Designation: log a x = b, where a is the base, x is the argument, b is what the logarithm is actually equal to.

For example, 2 3 = 8 ⇒ log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). With the same success log 2 64 = 6, since 2 6 = 64.

The operation of finding the logarithm of a number to a given base is called logarithmization. So, let's add a new line to our table:

2 1	2 2	2 3	2 4	2 5	2 6
2	4	8	16	32	64
log 2 2 = 1	log 2 4 = 2	log 2 8 = 3	log 2 16 = 4	log 2 32 = 5	log 2 64 = 6

Unfortunately, not all logarithms are calculated so easily. For example, try finding log 2 5 . The number 5 is not in the table, but logic dictates that the logarithm will lie somewhere on the segment. Because 2 2< 5 < 2 3 , а чем больше степень двойки, тем больше получится число.

Such numbers are called irrational: the numbers after the decimal point can be written ad infinitum, and they are never repeated. If the logarithm turns out to be irrational, it is better to leave it that way: log 2 5, log 3 8, log 5 100.

It is important to understand that a logarithm is an expression with two variables (the base and the argument). At first, many people confuse where the basis is and where the argument is. To avoid annoying misunderstandings, just look at the picture:

Before us is nothing more than the definition of a logarithm. Remember: logarithm is a power, into which the base must be built in order to obtain an argument. It is the base that is raised to a power - it is highlighted in red in the picture. It turns out that the base is always at the bottom! I tell my students this wonderful rule at the very first lesson - and no confusion arises.

We've figured out the definition - all that remains is to learn how to count logarithms, i.e. get rid of the "log" sign. To begin with, we note that two important facts follow from the definition:

1. The argument and the base must always be greater than zero. This follows from the definition of the degree rational indicator, to which the definition of a logarithm comes down.
2. The base must be different from one, since one to any degree still remains one. Because of this, the question “to what power must one be raised to get two” is meaningless. There is no such degree!

Such restrictions are called range of acceptable values(ODZ). It turns out that the ODZ of the logarithm looks like this: log a x = b ⇒ x > 0, a > 0, a ≠ 1.

Note that there are no restrictions on the number b (the value of the logarithm). For example, the logarithm may well be negative: log 2 0.5 = −1, because 0.5 = 2 −1.

However, now we are considering only numerical expressions, where it is not required to know the VA of the logarithm. All restrictions have already been taken into account by the authors of the tasks. But when logarithmic equations and inequalities come into play, DL requirements will become mandatory. After all, the basis and argument may contain very strong constructions that do not necessarily correspond to the above restrictions.

Now let's consider general scheme calculating logarithms. It consists of three steps:

1. Express the base a and the argument x as a power with the minimum possible base greater than one. Along the way, it’s better to get rid of decimals;
2. Solve the equation for variable b: x = a b ;
3. The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be visible already in the first step. The requirement that the base be more than one, is very relevant: it reduces the likelihood of error and greatly simplifies the calculations. It’s the same with decimal fractions: if you immediately convert them into ordinary ones, there will be many fewer errors.

Let's see how this scheme works using specific examples:

Task. Calculate the logarithm: log 5 25

1. Let's imagine the base and argument as a power of five: 5 = 5 1 ; 25 = 5 2 ;

Let's create and solve the equation:
log 5 25 = b ⇒ (5 1) b = 5 2 ⇒ 5 b = 5 2 ⇒ b = 2 ;

2. We received the answer: 2.

Task. Calculate the logarithm:

Task. Calculate the logarithm: log 4 64

1. Let's imagine the base and argument as a power of two: 4 = 2 2 ; 64 = 2 6 ;
2. Let's create and solve the equation:
   log 4 64 = b ⇒ (2 2) b = 2 6 ⇒ 2 2b = 2 6 ⇒ 2b = 6 ⇒ b = 3 ;
3. We received the answer: 3.

Task. Calculate the logarithm: log 16 1

1. Let's imagine the base and argument as a power of two: 16 = 2 4 ; 1 = 2 0 ;
2. Let's create and solve the equation:
   log 16 1 = b ⇒ (2 4) b = 2 0 ⇒ 2 4b = 2 0 ⇒ 4b = 0 ⇒ b = 0 ;
3. We received the answer: 0.

Task. Calculate the logarithm: log 7 14

1. Let's imagine the base and argument as a power of seven: 7 = 7 1 ; 14 cannot be represented as a power of seven, since 7 1< 14 < 7 2 ;
2. From the previous paragraph it follows that the logarithm does not count;
3. The answer is no change: log 7 14.

A small note on the last example. How can you be sure that a number is not an exact power of another number? It's very simple - just break it down into prime factors. If the expansion has at least two different factors, the number is not an exact power.

Task. Find out whether the numbers are exact powers: 8; 48; 81; 35; 14 .

8 = 2 · 2 · 2 = 2 3 - exact degree, because there is only one multiplier;
48 = 6 · 8 = 3 · 2 · 2 · 2 · 2 = 3 · 2 4 - is not an exact power, since there are two factors: 3 and 2;
81 = 9 · 9 = 3 · 3 · 3 · 3 = 3 4 - exact degree;
35 = 7 · 5 - again not an exact power;
14 = 7 · 2 - again not an exact degree;

Note also that the prime numbers themselves are always exact powers of themselves.

Decimal logarithm

Some logarithms are so common that they have a special name and symbol.

The decimal logarithm of x is the logarithm to base 10, i.e. The power to which the number 10 must be raised to obtain the number x. Designation: lg x.

For example, log 10 = 1; lg 100 = 2; lg 1000 = 3 - etc.

From now on, when a phrase like “Find lg 0.01” appears in a textbook, know that this is not a typo. This is a decimal logarithm. However, if you are unfamiliar with this notation, you can always rewrite it:
log x = log 10 x

Everything that is true for ordinary logarithms is also true for decimal logarithms.

Natural logarithm

There is another logarithm that has its own designation. In some ways, it's even more important than decimal. It's about about the natural logarithm.

The natural logarithm of x is the logarithm to base e, i.e. the power to which the number e must be raised to obtain the number x. Designation: ln x .

Many will ask: what is the number e? This is an irrational number; its exact value cannot be found and written down. I will give only the first figures:
e = 2.718281828459...

We won’t go into detail about what this number is and why it is needed. Just remember that e is the base of the natural logarithm:
ln x = log e x

Thus ln e = 1 ; ln e 2 = 2; ln e 16 = 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number irrational. Except, of course, for one: ln 1 = 0.

For natural logarithms, all the rules that are true for ordinary logarithms are valid.

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

Let's explain it more simply. For example, \(\log_(2)(8)\) is equal to the power to which \(2\) must be raised 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 logarithm

Any logarithm has the following “anatomy”:

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

How to calculate logarithm?

To calculate the logarithm, you need to answer the question: to what power 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 one. That's why:

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

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

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

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

d) To what power must \(\sqrt(7)\) be raised to obtain \(\sqrt(7)\)? Firstly, any number to the first power is equal to itself.

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

e) To what power must \(3\) be raised to obtain \(\sqrt(3)\)? From we know what it is fractional power, and that means Square root is the power of \(\frac(1)(2)\) .

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

Example : Calculate 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 a logarithm: \(\log_(a)(c)=b\) \(\Leftrightarrow\) \(a^(b)=c\)
\((4\sqrt(2))^(x)=8\)		What connects \(4\sqrt(2)\) 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 properties of the degree: \(a^(m)\cdot a^(n)=a^(m+n)\) and \((a^(m))^(n)=a^(m\cdot n)\)
\(2^(\frac(5)(2)x)=2^(3)\)		The bases are equal, we move on to 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 smartest ones will say: “X is a little less than two.” How exactly to write this number? To answer this question, the logarithm was invented. Thanks to him, the answer here can be written as \(x=\log_(3)(8)\).

I want to emphasize that \(\log_(3)(8)\), like any logarithm is just a number. Yes, it looks unusual, but it’s short. Because if we wanted to write it in the form decimal, 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 brought to the same base. This means you can’t do without a logarithm. Let's use the definition of logarithm: \(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)
\(\log_(4)(10)=5x-4\)		Let's flip the equation so that X is on the left
\(5x-4=\log_(4)(10)\)		Before us. Let's move \(4\) to the right. And don't be afraid of the logarithm, treat it like an ordinary number.
\(5x=\log_(4)(10)+4\)		Divide the equation by 5
\(x=\)\(\frac(\log_(4)(10)+4)(5)\)		This is our root. Yes, it looks unusual, but they don’t choose the answer.

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

Decimal and natural logarithms

As stated in the definition of a logarithm, its base can be any positive number except one \((a>0, a\neq1)\). And among all the possible bases, 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 Euler's number \(e\) (equal to approximately \(2.7182818…\)), and the logarithm is written as \(\ln(a)\).

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

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

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

Basic logarithmic identity

Logarithms have many properties. One of them is called the “Basic Logarithmic Identity” and looks like this:

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

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

Let us recall a short notation of the definition of logarithm:

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 other 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 instead of two you can write \(\log_(2)(4)\).

But \(\log_(3)(9)\) is also equal to \(2\), which means we can also write \(2=\log_(3)(9)\) . Likewise 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 two as a logarithm with any base anywhere (be it in an equation, in an expression, or in an inequality) - we simply write the base squared as an argument.

It’s the same with the 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 meaning of the expression \(\frac(\log_(2)(14))(1+\log_(2)(7))\)

Solution :

Answer : \(1\)

Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called main properties.

You definitely need to know these rules - without them not a single serious problem can be solved. logarithmic problem. In addition, there are very few of them - you can learn everything in one day. So let's get started.

Adding and subtracting logarithms

Consider two logarithms with the same bases: 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 logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Note: key moment Here - identical grounds. If the reasons are different, these rules do not work!

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

Log 6 4 + log 6 9.

Since logarithms have the same bases, 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.

As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many are built on this fact test papers. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.

Extracting the exponent from the logarithm

Now let's complicate the task a little. What if the base or argument of a logarithm is a power? 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 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 of the 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 logarithm sign 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 using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the meaning of the expression:

[Caption for the picture]

Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:

[Caption for the picture]

I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator contain 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 is what was done. The result was 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 reasons are different? What if they are not exact powers of the same number?

Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:

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

[Caption for the picture]

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

[Caption for the picture]

From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears 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 problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:

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

Note that the arguments of both logarithms contain exact powers. 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 “reverse” the second logarithm:

[Caption for the picture]

Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with 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 this down and get rid of the indicators:

[Caption for the picture]

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

[Caption for the picture]

Basic logarithmic identity

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

In the first case, the number n becomes an indicator of the degree standing in the argument. Number n can be absolutely anything, because it’s just a logarithm value.

The second formula is actually a paraphrased definition. That’s what it’s called: the basic logarithmic identity.

In fact, what will happen if the number b raise to such a power that the number b to this power gives the number a? That's right: you get this same number a. Read this paragraph carefully again - many people get stuck on it.

Like formulas for moving to a new base, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the meaning of the expression:

[Caption for the picture]

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

[Caption for the picture]

If anyone doesn't know, this was a real task from the Unified State Exam :)

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They constantly appear in problems and, surprisingly, create problems even for “advanced” students.

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