# 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".

Continues below ⇩

### Examples

1. Expand

log 7x

as the sum of 2 logarithms.

2. Using your calculator, show that

`log (20/5) = log\ 20 − log\ 5`.

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

**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)`

2. Express

2 log

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

as the logarithm of a single quantity.

**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`

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

log

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

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