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 xxlog_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.

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