# 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.

Exponents | Logarithms |
---|---|

`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 "log_{10}7".

### Examples

1. Expand

log 7x

as the sum of 2 logarithms.

Answer

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`.

Answer

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.

3. Express as a multiple of logarithms: log *x*^{5}.

Answer

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.

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

log

_{6 }6 = log_{10 }10 = log_{x}x= log_{a}a= 1

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

6

^{1}= 610

^{1}= 10

x^{1}=x

a^{1}=a

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

log

_{7 }1 = log_{10 }1 = log_{e}1 = log_{x}1 = 0

The equivalent statments in exponential form are:

7

^{0}= 110

^{0}= 1

e^{0}= 1

x^{0}= 1

### Exercises

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

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

Answer

`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)`

Please support IntMath!

2. Express

2 log

_{e }2 + 3 log_{e }n

as the logarithm of a single quantity.

Answer

Applying the logarithm laws, we have:

2 log

2 + 3 log_{e}_{e}n= log

4 + log_{e}_{e}n^{3}= log

4_{e}n^{3}

**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`

Answer

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`

Get the Daily Math Tweet!

IntMath on Twitter

**4. **Solve for *y* in terms of *x*:

log

_{2 }x+ log_{2 }y= 1

Answer

Using the first log law, we can write:

log

_{2}xy= 1

Then *xy* = 2^{1}

So

`y=2/x`

### Search IntMath, blog and Forum

### Online Math Solver

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

Go to: Online algebra solver

### Math Lessons on DVD

Math videos by MathTutorDVD.com

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!