Exponentiation w/ independent variable

Hello,

I have a quick question regarding solving a weird log.

Given the following: log (with base x-1) of x = f(x). Is it permissible to exponentiate in the following manner:

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

(x-1)^y = x and then take the derivative

y ln (x-1) = ln x

1/x-1 * y + y' * ln (x-1) = 1/x ... ect solving for y' is trivial.

I know the alternate (which IS correct) is to rewrite using change of base yielding

log x/ log x-1 = y and then taking the derivative. However, does the first method also arrive at the correct answer, or is it not legal?

Any help is appreciated,

Thanks.

[Primeinator]

I do not immediately see a problem by exponentiating in the fashion above- i.e. using (x-1) as a base to cancel the logarithm and then taking the log of both sides to bring down your dependent variable y looks sound. In addition, to my knowledge there is no problem with using change of base formula when the independent variable is in the base of the log. However, I am just an amateur on these forums. Perhaps someone with more expertise can speak on this subject?

[Primeinator]

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.

[davieddy]

If y=u/v then yv = u

Differentiate both equations and solve for y'.

[davieddy]

BTW what is x "independent" of?