Logarithm Laws Made Easy: A Complete Guide with Examples

Logarithm Laws: Video Lesson

What are the Laws of Logarithms?

The laws of logarithms are algebraic rules that allow for the simplification and rearrangement of logarithmic expressions. The 3 main logarithm laws are:
  1. The Product Law: log(mn) = log(m) + log(n).
  2. The Quotient Law: log(m/n) = log(m) – log(n).
  3. The Power Law: log(mk) = k·log(m).

The three fundamental laws of logarithms are shown below.

the laws of logarithms

Rules when using the Laws of Logarithms

The rules for the laws of logarithms are:

  • To simplify logarithms with the product or quotient laws, the bases of the logarithms must be equal.
  • The base of the logarithms must be greater than zero.
  • The base of the logarithms cannot be equal to 1.
  • The inputs of the logarithms must be greater than zero.
  • The logarithm of 1 is equal to zero, no matter what the base of the logarithm is.

The three fundamental laws of logarithms can also be written as:

  • loga(m) + loga(n) = loga(mn)
  • loga(m) – loga(n) = loga(m/n)
  • loga(mk) = kloga(m)
3 basic logarithm laws

Examples of Using Logarithm Laws

The three most important logarithm laws are shown below with an example of each.

example of using the addition log law
The product logarithm law:

When adding logarithms with the same base, the inputs to the logarithms can be multiplied.

The 4 and 5 inside the logarithms are multiplied to obtain 20.

example of using the subtraction log law
The quotient logarithm law:

When subtracting logarithms with the same base, the first logarithm is divided by the logarithm being subtracted.

10 is divided by 5 inside the logarithm to obtain log(2).

example of using the power log law
The power logarithm law:

The coefficient of the logarithm can be moved to become the power of the input to the logarithm.

5 is raised to the power of 2 to obtain 25.

List of the Logarithm Laws

The following table shows the complete list of log laws with examples of how they are used:

Logarithm LawFormulaExample
Product Lawloga(m) + loga(n) = loga(mn)log(2) + log(3) = log(6)
Quotient Lawloga(m) – loga(n) = loga(m/n)log(12) – log(3) = log(4)
Power Lawloga(mk) = k ·loga(m)log(9) = log(32) = 2log(3)
Inverse Logarithm Propertyloga(ak) = klog2(8) = log2(23) = 3
Zero Lawloga(1) = 0log3(1) = 0
Logarithm of a Reciprocalloga(1/m) = -loga(m)log(1/2) = -log(2)
Identity Property of a Logarithmloga(a) = 1log5(5) = 1
Inverse Property of an Exponentaloga(k) = k3log3(5) = 5
Change of Base Lawloga(m) = logc(m)/logc(a)log2(5) = log3(5)/log3(2)

Here is a visual list of 5 of the most commonly used logarithm laws.

The colours used in this list show how the different rearrangements are made in a more simple, visual manner.

list of logarithm laws

The following sections take a look at each of the logarithm laws in greater detail.

The logarithm laws are:

  1. The Product Rule
  2. The Quotient Rule
  3. The Power Rule
  4. The Inverse Property of Logarithms
  5. The Zero Rule
  6. The Identity Rule
  7. The Inverse Property of the Exponent

The Product (Addition) Logarithm Law

To add two or more logarithms that have the same base, simply multiply the numbers inside the logarithms. For example, log(3)+log(2) = log(6). The result is a single logarithm with the same base as those being added.

How to add logarithms

The formula for the product law of logarithms is given as:

the log base a of m plus the log base a of n equals log base a m n

The product rule of logarithm laws

Two logarithms can only be simplified using the addition (product) rule if they have the same base. This base remains the same in the resulting answer.

For example, log3(2) + log3(5) = log3(2 × 5) = log3(10).

addition product logarithm law

Here is another example of adding logarithms using log laws and then simplifying.

log4(2) + log4(8) = log4(16)

Since two logarithms of the same base were added, the resulting logarithm input was equal to the product of the inputs given.

Now log4(16) = 2 because 42 = 16.

A logarithm simply asks what power the base of the logarithm must be raised to in order to obtain the input.

4 must be raised to the power of 2 to obtain 16 and so, the final answer of this question is just 2.

multiplication log law

The product rule for logarithms can also be written in reverse using the formula:

log base a m n equals the log base a of m plus the log base a of n

The product rule of logarithm laws

The product rule of logarithms states that a single logarithm can be separated into the sum of individual logarithms which have inputs that multiply to make the input of the original logarithm.

For example, log(21) = log(3) + log(7).

Here are some examples of how to add logarithms using the addition log law.

QuestionSolution
log(5) + log(3)= log(5×3)= log(15)
log3(2) + log3(7)= log3(2×7)= log3(14)
log(2) + log(3) + log(4)= log(2×3×4)= log(24)
log(1) + log(20) + log(0.25)= log(1×20×0.25)= log(5)
ln(4) + ln(5)= ln(4×5)= ln(20)
log(𝑥2) + log(y)= log(𝑥2 × y)= log(𝑥2y)

The Quotient (Subtraction )Logarithm Law

Logarithms of the same base can be subtracted by dividing their inputs. For example, log(8) – log(4) = log(2), since 8 ÷ 4 = 2. The result is a single logarithm with the same base as those being subtracted.

how to subtract logarithms

The formula for subtracting logarithms using the quotient law is given as:

the log base a of m minus the log base a of n equals the log base a of open paren m over n close paren

The quotient law of logarithms

Two or more logarithms can only be simplified using the quotient law if they have the same base. The input of the first logarithm is divided by the input of the logarithm being subtracted. The result is a single logarithm with the same base.

For example, log3(20) – log3(4) = log3(5).

The result is a single logarithm, also base 3.

subtraction quotient logarithm law

Here is another example of subtracting logarithms using the quotient logarithm law.

log5(100) – log5(4) = log3(100 ÷ 4) which equals log5(25).

log5(25) can be evaluated since 25 is a power of 5.

52 = 25 and so, log5(25) = 2.

division log law

The quotient rule for logarithms can also be written in reverse using the formula:

the log base a of open paren m over n close paren equals the log base a of m minus the log base a of n

The product rule of logarithm laws

The quotient rule of logarithms states that a single logarithm can be written as a logarithm subtracted from another logarithm.

For example, log(2/5) = log(2) – log(5).

Here are some examples of how to subtract logarithms using the subtraction log law.

QuestionSolution
log(10) – log(2)= log(10÷2)= log(5)
log3(6) – log3(3)= log3(6÷3)= log3(2)
log(20) – log(2) – log(5)= log(20÷2÷5)= log(2)
log(30) – log(5) – log(1)= log(30÷5÷1)= log(6)
ln(12) – ln(6)= ln(12÷6)= ln(2)
log(𝑥2) – log(y)= log(𝑥2 ÷ y)= log(𝑥2/y)

The Logarithm Law for Exponents

If there is a coefficient in front of a logarithm, the power law of logarithms states that the input to the logarithm can be raised to the power of this coefficient. For example, 2log(3) = log(32) which equals log(9).

power logarithm law

The formula for the power law of logarithms is:

m the log base a of b equals the log base a of open paren b to the power of m close paren

The power law of logarithms

For example, 3log2(10) can be written as log2(103).

This is because the coefficient of 3 immediately in front of the log can be moved so that it is acting as a power on the number inside the log.

Now log2(103) = log2(1000).

how to do logarithms with a coefficient

Reciprocal Rule of Logarithms

When the reciprocal of the input to a logarithm is taken, the result is equal to -1 multiplied by the logarithm of the original input. That is log(1/b) = -log(b). For example, log(1/2) = -log(2).

We can see that since 1 over b equals b to the negative 1 power then the log of open paren 1 over b close paren equals the log of open paren b to the negative 1 power close paren.

We can use the power law of logarithms to move the power of -1 down to become the coefficient of the logarithm so that the log of open paren b to the negative 1 power close paren equals negative 1 times log b or more simply, the log of open paren b to the negative 1 power close paren equals negative log b.

reciprocal rule of logarithms

For example, log one half equals negative log 2.

example of the reciprocal logarithm law

The formula for the reciprocal of the input to a logarithm is given by:

the log of open paren 1 over b close paren equals negative log b

Reciprocal of the input to a logarithm

The reciprocal logarithm law can be applied in the following circumstances.

  • log one tenth equals negative log 10
  • l n of open paren e over 5 close paren equals negative l n of open paren 5 over e close paren
  • log 1 equals negative log 1 (since both sides of the equation equal zero)
  • log 2 equals negative log one half

Logarithms Involving a Square Root

When the coefficient of a logarithm is equal to one half, the input inside the logarithm can be raised to the power of this half. Raising a value to the power of one half is equivalent to finding its square root. Therefore (1/2)loga(b) = loga(√b). For example, (1/2)log(16) = log(√16) = log(4).

The formula for the a logarithm with a coefficient of one half is given as:

one half the log base a of b equals the log base a of open paren the square root of b close paren

Formula for a logarithm with a coefficient of one half

In the example below, using the power logarithm law, (1/2)log(16) = log(161/2).

When 16 is raised to the power of one half, it is equivalent of finding the square root of 16.

Therefore (1/2)log(16) = log(161/2) = log(√16) = log(4).

Log laws with a square root

In general, b raised to the 1 over n power equals to the power of n the square root of b.

And so, the formula for any fractional coefficient of a logarithm is:

1 over n the log base a of b equals the log base a of open paren the n-th root of b close paren

Formula for a fractional coefficient of a logarithm

For example, one third the log base 5 of 8 equals the log base 5 of open paren the cube root of 8 close paren and so, one third the log base 5 of 8 equals the log base 5 of 2.

The Inverse Property of Logarithms

The inverse property of logarithms states that loga(ak) = k. When the base number inside the logarithm is equal to the base of the logarithm, the result is simply the value of the exponent inside the logarithm. For example, log3(35) = 5.

The formula for the inverse property of logarithms is:

the log base a of open paren a to the power of k close paren equals k

The Inverse Property of Logarithms

The other example of the inverse property of logarithms listed above is the log base 2 of open paren 2 to the power of x close paren equals x.

That is because when 2 is raised to the power of 𝑥, we obtain 2𝑥.

Logarithm Law: The Zero Rule

The logarithm of one is equal to zero no matter what the base of the logarithm is. That is, loga(1) = 0 for all valid values of ‘a’. Some of the many examples include: log2(1) = 0, log5(1) = 0, log(1) = 0 and ln(1) = 0.

This rule is true since the value of a logarithmic expression is always equal to the power that the base of the logarithm must be raised to in order to obtain the value of the input of the logarithm.

Since any positive or negative number raised to the power of zero is equal to 1 (that is, a0=1), then loga(1) must be equal to 0.

logarithm zero rule

Logarithm Law: The Identity Rule

The identity rule of logarithms states that if the input to a logarithm is equal in value to the base of the logarithm, the result is equal to 1. That is, loga(a) = 1. For example, log3(3) = 1 and log2(2) = 1.

identity rule of logarithms

This rule works since a logarithmic expression is equal in value to the power that its base must be raised to in order to obtain the value which is the input to the logarithm.

Since a1 = a, then loga(a) must equal 1.

The identity rule of logarithms is applied to natural logarithms in the sense that ln(e) = 1. This is because ln(e) means loge(e).

Logarithm Law: Inverse Property of Exponents

The inverse property of exponents states that if a given number is raised to the power of a logarithm which has the same base as this given number, then the result is simply equal to the input of the logarithm. That is, aloga(k) = k. For example, 2log2(5) = 5.

inverse property of exponents and logarithms

The inverse property of exponents can be applied to natural logarithms in the following manner:

eln(k) = k

This is true since ln(k) means loge(k).

For example, eln(2) = 2.

Logarithm Laws: Examples and Solutions

Here are some examples with solutions of using logarithm laws to simplify and expand expressions.

It is important to understand that the addition and subtraction logarithm laws can only be used when we have one logarithm added to or subtracted from another logarithm and there are no coefficients in front of either logarithm.

Therefore, it is important to always use the power logarithm law to move any coefficients from in front of the logarithms before adding or subtracting them.

For example, use logarithm laws to simplify 2 log 2 plus 3 log 10.

Both logarithms have coefficients and so, we use the power log law so that 2 log 2 equals the log of open paren 2 squared close paren and 3 log 10 equals the log of open paren 10 cubed close paren.

Therefore we know that 2 log 2 equals log 4 and 3 log 10 equals log 1000.

Therefore 2 log 2 plus 3 log 10 becomes log 4 plus log 1000.

Now that the coefficients have been moved and we simply have one logarithm plus another logarithm, we can add them using the addition (product) law of logarithms.

We simply multiply the inputs of 4 and 1000.

log 4 plus log 1000 equals log 4000.

simplifying expression using logarithm laws

Here is another example of simplifying a logarithmic expression using logarithm laws.

Simplify 3 the log base 5 of 2 minus one half the log base 5 of 4.

We move the coefficients first so that they become the powers of the inputs to the logarithm.

In the first logarithm: 3 the log base 5 of 2 equals the log base 5 of open paren 2 cubed close paren equals the log base 5 of 8

and in the second logarithm: one half the log base 5 of 4 equals the log base 5 of open paren 4 raised to the one half power close paren equals the log base 5 of open paren the square root of 4 close paren equals the log base 5 of 2.

Therefore 3 the log base 5 of 2 minus one half the log base 5 of 4 becomes the log base 5 of 8 minus the log base 5 of 2.

Since the logarithms have the same base, we can subtract them using the subtraction (quotient) law.

We simply divide 8 by 2 to obtain 4.

the log base 5 of 8 minus the log base 5 of 2 equals the log base 5 of 4.

simplify using log laws example

Here are some further examples of using logarithm laws:


  • Simplify the log base 3 of 5 plus the log base 3 of 4 minus the log base 3 of 2.

All logarithms are the same base and so we multiply the numbers in the logarithms that are being added and then divide by the number inside the logarithm being subtracted.

the log base 3 of 5 plus the log base 3 of 4 minus the log base 3 of 2 equals the log base 3 of open paren 5 times 4 divided by 2 close paren

Therefore: the log base 3 of 5 plus the log base 3 of 4 minus the log base 3 of 2 equals the log base 3 of 10


  • Simplify the log base 2 of 16 plus the log base 2 of 2.

Both 16 and 2 are powers of 2, which is the base of the logarithms.

We can write the log base 2 of 16 as the log base 2 of open paren 2 to the fourth power close paren which is equal to 4.

the log base 2 of 2 is just equal to 1.

Therefore, the log base 2 of 16 plus the log base 2 of 2 equals 4 plus 1 equals 5 and so, the answer is 5.

Alternatively, we can use the addition (product) law immediately so that the log base 2 of 16 plus the log base 2 of 2 equals the log base 2 of 32.

Then we can use the fact that 25=32 to evaluate the log base 2 of 32 equals 5.


  • Evaluate log 200 minus log 2.

We can use the subtraction (quotient) law so that log 200 minus log 2 equals log 100.

Using the fact that log(100) means log10(100) and also the fact that 102=100, we can evaluate log(100) = 2.

Therefore log 200 minus log 2 equals 2.


  • Expand the logarithmic expression of the log base 5 of open paren 25 x cubed y squared z close paren.

Since the 25, x cubed, y squared and z are multiplied together inside the logarithm, we can writethe log base 5 of open paren 25 x cubed y to the power of y z close paren equals the log base 5 of 25 plus the log base 5 of open paren x cubed close paren plus the log base 5 of open paren y squared close paren plus the log base 5 of z.

We can see that log5(25) = 2 since 52 = 25.

Therefore the log base 5 of open paren 25 x cubed y to the power of y z close paren equals 2 plus the log base 5 of open paren x cubed close paren plus the log base 5 of open paren y squared close paren plus the log base 5 of z.

We can then use the power law to bring the coefficients down in front of the logarithms to obtain:

the log base 5 of open paren 25 x cubed y to the power of y z close paren equals 2 plus 3 the log base 5 of x plus 2 the log base 5 of y plus the log base 5 of z.


  • Expand the logarithmic expression the log base 2 of open paren the fraction with numerator 6 x and denominator y to the fifth power close paren.

All terms on the top of the fraction are multiplied together and so, can be written as the summation of individual logs. The term on the bottom of the fraction are subtracted as we are dividing by this.

We write the log base 2 of open paren the fraction with numerator 6 x and denominator y to the fifth power close paren equals the log base 2 of 6 plus the log base 2 of x minus the log base 2 of open paren y to the fifth power close paren.

We can write log2(6) as log2(2) + log2(3).

We can therefore write log2(6) as 1 + log2(3).

Therefore the log base 2 of open paren the fraction with numerator 6 x and denominator y to the fifth power close paren equals 1 plus the log base 2 of 3 plus the log base 2 of x minus the log base 2 of open paren y to the fifth power close paren.

Finally, we can bring down the power of 5 on the final logarithm using the power law.

the log base 2 of open paren the fraction with numerator 6 x and denominator y to the fifth power close paren equals 1 plus the log base 2 of 3 plus the log base 2 of x minus 5 the log base 2 of y.


  • Expand the logarithmic expression of l n of open paren the fraction with numerator 2 and denominator the square root of e close paren.

We can write l n of open paren the fraction with numerator 2 and denominator the square root of e close paren equals l n of 2 minus l n of open paren the square root of e close paren.

We can then write this as l n of open paren the fraction with numerator 2 and denominator the square root of e close paren equals l n of 2 minus l n of open paren e raised to the one half power close paren.

We bring down the power using the power law so that l n of open paren the fraction with numerator 2 and denominator the square root of e close paren equals l n of 2 minus one half l n of e.

Finally, we use the fact that ln(e) = 1 so that:

l n of open paren the fraction with numerator 2 and denominator the square root of e close paren equals l n of 2 minus one half.

Expanding Logarithms Using Logarithm Laws

Single logarithms can be expanded into multiple logarithms of the same base using logarithm laws.

For example, using the addition (product) law, we can write log(20) as log(20) = log(5) + log(4).

Then log(4) can also be written as log(22), which can be written as 2log(2).

Therefore log(20) can be written as log(20) = log(5) + 2log(2).

expanding a logarithm using log laws

Here is another example of expanding a logarithmic expression using logarithm laws.

Using the addition and subtraction logarithm laws, we can write the log of open paren the fraction with numerator 10 x cubed and denominator y to the fifth power close paren equals log 10 plus the log of open paren x cubed close paren minus the log of open paren y to the fifth power close paren.

All terms on top of the fraction inside the logarithm are positive and the log(y5) is subtracted since it has been divided inside the logarithm.

We can then use the power law of logarithms to bring down the powers as coefficients.

We can also evaluate log(10) = 1.

log 10 plus the log of open paren x cubed close paren minus the log of open paren y to the fifth power close paren equals 1 plus 3 log x minus 5 log y

Therefore the log of open paren the fraction with numerator 10 x cubed and denominator y to the fifth power close paren equals 1 plus 3 log x minus 5 log y.

expanding a logarithmic expression

Change of Base Rule

The change of base formula for logarithms is loga(b) = logc(b) ÷ logc(a). For example log2(10) = log(10) ÷ log(2).

The change of base logarithm formula is:

the log base a of b equals the log base c of b over the log base c of a

The change of base logarithm formula

To divide logarithms that have the same base, the change of base formula can be used. That is, logc(b) ÷ logc(a) = loga(b). For example, log2(27) ÷ log2(3) = log3(27) and this can be evaluated since log3(27) = 3.

Proof of the Logarithm Laws

The following is the algebraic proof of the product (addition) law of logarithms.

  • Let x equals a to the power of m and take logarithms of both sides to obtain log x equals the log of open paren a to the power of m close paren. Then use the power law of logarithms to bring the power of m down as the coefficient of the logarithm. That is, log x equals the log of open paren a to the power of m close paren equals m log a.
  • Take y equals a to the power of n and follow the same process as above to obtain log y equals the log of open paren a to the power of n close paren.
  • Multiply x equals a to the power of m and y equals a to the power of n to obtain x y equals a to the power of m times a to the power of n which is more simply written as x y equals a raised to the m plus n power.
  • Take logarithms of both sides to obtain log x y equals the log of open paren a raised to the m plus n power close paren. Then use the power law of logarithms to bring the (m+n) down as the coefficient of the log so that log x y equals the log of open paren a raised to the m plus n power close paren equals open paren m plus n close paren times log a.
  • Then expand the brackets to obtain log x y equals m log a plus n log a.
  • Finally substitute log x equals m log a and log y equals n log a from the previous calculations so that log x y equals m log a plus n log a becomes log x y equals log x plus log y.

proof of the product logarithm law

The following is the algebraic proof of the quotient (subtraction) law of logarithms:

  • Let x equals a to the power of m and take logarithms of both sides to obtain log x equals the log of open paren a to the power of m close paren. Then use the power law of logarithms to bring the power of m down as the coefficient of the logarithm. That is, log x equals the log of open paren a to the power of m close paren equals m log a.
  • Take y equals a to the power of n and follow the same process as above to obtain log y equals the log of open paren a to the power of n close paren.
  • Divide 𝑥 by y so that x over y equals a to the power of m divided by a to the power of n and so, x over y equals a raised to the m minus n power.
  • Take logarithms of both sides so that the log of open paren x over y close paren equals the log of open paren a raised to the m minus n power close paren.
  • This can be written using the power law of logarithms as the log of open paren x over y close paren equals open paren m minus n close paren times log a.
  • Expanding the brackets, the log of open paren x over y close paren equals m log a minus n log a.
  • Finally substitute log x equals m log a and log y equals n log a from the previous calculations so that the log of open paren x over y close paren equals m log a minus n log a becomes the log of open paren x over y close paren equals log x minus log y.
proof of the division law of logarithms

The following is algebraic proof of the exponent (power) law of logarithms:

  • Consider the expoenential equation of x equals a to the power of m and write this in logarithmic form as the log base a of x equals m.
  • Returning to x equals a to the power of m and raise both sides to the power of k to obtain x to the power of k equals a to the power of m to the power of k.
  • This can be written more simply as x to the power of k equals a raised to the m k power.
  • Now write x to the power of k equals a raised to the m k power in logarithmic form as the log base a of open paren x to the power of k close paren equals m k.
  • Now substitute the log base a of x equals m as previously found into the log base a of open paren x to the power of k close paren equals m k in order to obtain the log base a of open paren x to the power of k close paren equals the log base a of x times k.
  • This can be written in the more familiar format as the log base a of open paren x to the power of k close paren equals k the log base a of x.
proof of the exponent law of logarithms