## Log rules

### May 19, 2016

Why $log_a x^y$ = $y log_a x$, y $\in$ Q, x > 0, a > 0  ?

i) Show $log_a x^n$${n} log_a x$ , n $\in \mathbb{N}$, x > 0, a > 0

Consider: $log_a x^2$ = $log_a xx$$log_a x + log_a x$  = $2log_a x$

Similarly, and inductively:

$log_a x^n$ = $log_a \underbrace{xxx ... x}_\text{n terms }$$\underbrace{log_a x + log_a x + ... + log_a x}_\text{n terms }$  = $nlog_a x$ as required.

Now for rational powers

ii) Show $log_a x^q$${q} log_a x$ , q $\in \mathbb{Q}$, x > 0, a > 0

Consider: $log_a x$ = $log_a x^1$= $log_a( {x^\frac{1}{n}})^n$ = $n log_a( {x^\frac{1}{n}})$ From (i) above and still n $\in$ N.

Therefore: $log_a x$ = $n log_a( {x^\frac{1}{n}})$

Hence: $\frac{1}{n} log_a x$ = $log_a( {x^\frac{1}{n}})$  as required.

Similarly: $log_a( {x^\frac{m}{n}}) = log_a( {x^\frac{1}{n}})^m = m log_a( {x^\frac{1}{n}}) = \frac{m}{n} log_a( {x})$  as required.