How to convert a logarithm to a common base. Properties of logarithms and examples of their solutions

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

Let's explain it easier. For example, \(\log_(2)(8)\) equal to the degree, 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 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. So:

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

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

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, and so the square root is the power of \(\frac(1)(2)\) .

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

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

Decision :

\(\log_(4\sqrt(2))(8)=x\)		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 equals x? 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 \(4^(5x-4)=10\)

Decision :

\(4^(5x-4)=10\)		\(4^(5x-4)\) 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 for the unit \((a>0, a\neq1)\). 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 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 "Main logarithmic identity' and looks like this:

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

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

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 \(36^(\log_(6)(5))\)

Decision :

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 \(\frac(\log_(2)(14))(1+\log_(2)(7))\)

Decision :

Answer : \(1\)

(from the Greek λόγος - "word", "relation" and ἀριθμός - "number") numbers b by reason a(log α b) is called such a number c, and b= a c, that is, log α b=c and b=ac are equivalent. The logarithm makes sense if a > 0, a ≠ 1, b > 0.

In other words logarithm numbers b by reason a formulated as an exponent to which a number must be raised a to get the number b(the logarithm exists only for positive numbers).

From this formulation it follows that the calculation x= log α b, is equivalent to solving the equation a x =b.

For example:

log 2 8 = 3 because 8=2 3 .

We note that the indicated formulation of the logarithm makes it possible to immediately determine logarithm value when the number under the sign of the logarithm is a certain power of the base. Indeed, the formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b by reason a equals with. It is also clear that the topic of logarithm is closely related to the topic degree of number.

The calculation of the logarithm is referred to logarithm. The logarithm is mathematical operation taking the logarithm. When taking a logarithm, the products of factors are transformed into sums of terms.

Potentiation is the mathematical operation inverse to logarithm. When potentiating, the given base is raised to the power of the expression on which the potentiation is performed. In this case, the sums of terms are transformed into the product of factors.

Quite often, real logarithms with bases 2 (binary) are used, e Euler number e ≈ 2.718 ( natural logarithm) and 10 (decimal).

On the this stage appropriate to consider samples of logarithms log 7 2 , ln √ 5, lg0.0001.

And the entries lg (-3), log -3 3.2, log -1 -4.3 do not make sense, since in the first of them a negative number is placed under the sign of the logarithm, in the second - a negative number in the base, and in the third - and a negative number under the sign of the logarithm and a unit in the base.

Conditions for determining the logarithm.

It is worth considering separately the conditions a > 0, a ≠ 1, b > 0. definition of a logarithm. Let's consider why these restrictions are taken. This will help us with an equality of the form x = log α b, called the basic logarithmic identity, which directly follows from the definition of the logarithm given above.

Take the condition a≠1. Since one is equal to one to any power, then the equality x=log α b can only exist when b=1, but log 1 1 will be any real number. To eliminate this ambiguity, we take a≠1.

Let us prove the necessity of the condition a>0. At a=0 according to the formulation of the logarithm, can only exist when b=0. And then accordingly log 0 0 can be any non-zero real number, since zero to any non-zero power is zero. To eliminate this ambiguity, the condition a≠0. And when a<0 we would have to reject the analysis of rational and irrational values of the logarithm, since the exponent with rational and irrational exponent is defined only for non-negative bases. It is for this reason that the condition a>0.

And the last condition b>0 follows from the inequality a>0, since x=log α b, and the value of the degree with a positive base a always positive.

Features of logarithms.

Logarithms characterized by distinctive features, which led to their widespread use to greatly facilitate painstaking calculations. In the transition "to the world of logarithms", multiplication is transformed into a much easier addition, division into subtraction, and raising to a power and taking a root are transformed into multiplication and division by an exponent, respectively.

The formulation of logarithms and a table of their values (for trigonometric functions) was first published in 1614 by the Scottish mathematician John Napier. Logarithmic tables, enlarged and detailed by other scientists, were widely used in scientific and engineering calculations, and remained relevant until electronic calculators and computers began to be used.

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 both the exact value of the exponent 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 - without them, not a single serious logarithmic problem. 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 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 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 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:

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 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.

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

The natural logarithm is the logarithm whose basis is 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 .

By the look 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:

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:

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.

Instruction

Write down the given logarithmic expression. If the expression uses the logarithm of 10, then its notation is shortened and looks like this: lg b is decimal logarithm. If the logarithm has the number e as the base, then the expression is written: ln b is the natural logarithm. It is understood that the result of any is the power to which the base number must be raised to get the number b.

When finding the sum of two functions, you just need to differentiate them one by one, and add the results: (u+v)" = u"+v";

When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function, multiplied by the first function: (u*v)" = u"*v+v"*u;

In order to find the derivative of the quotient of two functions, it is necessary, from the product of the derivative of the dividend multiplied by the divisor function, to subtract the product of the derivative of the divisor multiplied by the divisor function, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;

If given complex function, then it is necessary to multiply the derivative of the inner function and the derivative of the outer one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).

Using the obtained above, you can differentiate almost any function. So let's look at a few examples:

y=x^4, y"=4*x^(4-1)=4*x^3;

y=2*x^3*(e^x-x^2+6), y"=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2 *x));
There are also tasks for calculating the derivative at a point. Let the function y=e^(x^2+6x+5) be given, you need to find the value of the function at the point x=1.
1) Find the derivative of the function: y"=e^(x^2-6x+5)*(2*x +6).

2) Calculate the value of the function in given point y"(1)=8*e^0=8

General principles of solution

Repeat the textbook mathematical analysis or higher mathematics, which is a definite integral. As you know, the solution definite integral there is a function whose derivative will give an integrand. This function is called primitive. According to this principle, the basic integrals are constructed.
Determine by the form of the integrand which of the table integrals fits in this case. It is not always possible to determine this immediately. Often, the tabular form becomes noticeable only after several transformations to simplify the integrand.

Variable substitution method

If the integrand is trigonometric function, whose argument is some polynomial, then try using the variable substitution method. To do this, replace the polynomial in the argument of the integrand with some new variable. Based on the ratio between the new and old variable, determine the new limits of integration. By differentiating this expression, find a new differential in . Thus you will receive the new kind the former integral, close or even corresponding to any tabular one.

Solution of integrals of the second kind

If the integral is an integral of the second kind, the vector form of the integrand, then you will need to use the rules for moving from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss ratio. This law allows you to go from the flow of the rotor to some vector function to the triple integral over the divergence of the given vector field.

Substitution of limits of integration

After finding the antiderivative, it is necessary to substitute the limits of integration. Plug in the value first upper limit into the expression for the antiderivative. You will receive some number. Next, subtract from the resulting number another number, the resulting lower limit to the antiderivative. If one of the integration limits is infinity, then substituting it into antiderivative function it is necessary to go to the limit and find what the expression tends to.
If the integral is two-dimensional or three-dimensional, then you will have to represent the geometric limits of integration in order to understand how to calculate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that limit the volume to be integrated.