# 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.

ExponentsLogarithms
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

### Examples

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.

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.

LHS

= log (20/5)

= log 4

= 0.60206 (using calculator)

Now

RHS

= log 20 − log 5

= 1.30103 − 0.69897 (using calculator)

= 0.60206

= LHS

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

Get the Daily Math Tweet!

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.

Easy to understand math videos:
MathTutorDVD.com

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

### Exercises

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

log_3((root(3)y)/8)

log_3((root(3)y)/8)

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

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

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

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

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:

log_3root(4)27

Let

log_3root(4)27=x

Then

3^x=root(4)27

Now

root(4)27=root(4)(3^3)=3^(3//4)

So x = 3/4.

Therefore

log_3root(4)27=3/4

4. Solve for y in terms of x:

log2 x + log2 y = 1

y=2/x