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Old 2010-07-11, 22:27   #3
Primeinator's Avatar
Feb 2005
Somewhere near M52..

3·5·61 Posts

Using the change of base formula:

f(x) = log(x) / log(x-1)

Via the quotient rule:

f ' (x) =[ log(x-1) / x - log(x) / (x-1) ] / (log(x-1))^2

Using what you mention:

f(x) = y = log(base x-1) of x

(x-1)^y = (x-1)^(log base x-1 of x) yielding

(x-1)^y = x Now taking the log of both sides:

y log (x-1) = log (x)

y = log (x) / log (x-1)

If you differentiate this expression, you will end up with an identical result to the above method. Therefore, I will venture to guess that exponentiating in this fashion is perfectly fine (at least in this scenario). Whether it is true in all scenarios or not... is a question that I cannot answer.

Last fiddled with by Primeinator on 2010-07-11 at 22:28
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