Skip to main content

3. The Logarithm Laws

by M. Bourne

Since a logarithm is simply an exponent which is just being written down on the line, we expect the logarithm laws to work the same as the rules for exponents, and luckily, they do.

`b^m × b^n = ` `b^(m+n)` `log_b xy = ` ` log_b x + log_b y`
`b^m ÷ b^n = ` `b^(m-n)` `log_b (x/y) = ` `log_b x − log_b y`
`(b^m)^n = b^(mn)` `log_b (x^n) =` ` n log_b x`
`b^1 = b` `log_b (b) = 1`
`b^0 = 1` `log_b (1) = 0`

Note: On our calculators, "log" (without any base) is taken to mean "log base 10". So, for example "log 7" means "log107".

Continues below


1. Expand

log 7x

as the sum of 2 logarithms.


Using the first law given above, our answer is

`log 7x = log 7 + log x`

Note 1: This has the same meaning as `10^7 xx 10^x = 10^(7+x)`

Note 2: This question is not the same as `log_7 x`, which means "log of x to the base `7`", which is quite different.

Easy to understand math videos:

2. Using your calculator, show that

`log (20/5) = log 20 − log 5`.


I am using numbers this time so you can convince yourself that the log law works.


`= log (20/5)`

`= log 4 `

`= 0.60206` (using calculator)



`= log 20 − log 5`

`= 1.30103 − 0.69897` (using calculator)

`= 0.60206 `


We have shown that the second logaritm law above works for our number example.

3. Express as a multiple of logarithms: log x5.


Using the third logarithm law, we have

`log x^5 = 5 log x`

We have expressed it as a multiple of a logarithm, and it no longer involves an exponent.

Please support IntMath!

Note 1: Each of the following is equal to 1:

log6 6 = log10 10 = logx x = loga a = 1

The equivalent statements, using ordinary exponents, are as follows:

61 = 6

101 = 10

x1 = x

a1 = a

Note 2: All of the following are equivalent to `0`:

log7 1 = log10 1 = loge1 = logx 1 = 0

The equivalent statments in exponential form are:

70 = 1

100 = 1

e0 = 1

x0 = 1


1. Express as a sum, difference, or multiple of logarithms:




` =log_3(root(3)y)-log_3(8)`


`=1/3log_3(y)-3 log_3(2)`

Get the Daily Math Tweet!
IntMath on Twitter

2. Express

2 loge 2 + 3 loge n

as the logarithm of a single quantity.


Applying the logarithm laws, we have:

2 loge 2 + 3 loge n

= loge 4 + loge n3

= loge 4n3

Get the Daily Math Tweet!
IntMath on Twitter

Note: The logarithm to base e is a very important logarithm. You will meet it first in Natural Logs (Base e) and will see it throughout the calculus chapters later.

3. Determine the exact value of:









So `x = 3/4`.



4. Solve for y in terms of x:

log2 x + log2 y = 1


Using the first log law, we can write:

log2 xy = 1

Then xy = 21



Please support IntMath!


Search IntMath, blog and Forum

Search IntMath

Online Math Solver

This math solver can solve a wide range of math problems.

Math Lessons on DVD

Math videos by

Easy to understand math lessons on DVD. See samples before you commit.

More info: Math videos

The IntMath Newsletter

Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!

See the Interactive Mathematics spam guarantee.