Exponentiation w/ independent variable
Hello,
I have a quick question regarding solving a weird log. Given the following: log (with base x1) of x = f(x). Is it permissible to exponentiate in the following manner: (x1)^y = (x1)^(log base (x1) of x) = (x1)^y = x and then take the derivative y ln (x1) = ln x 1/x1 * y + y' * ln (x1) = 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 x1 = 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. 
I do not immediately see a problem by exponentiating in the fashion above i.e. using (x1) 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?

Using the change of base formula:
f(x) = log(x) / log(x1) Via the quotient rule: f ' (x) =[ log(x1) / x  log(x) / (x1) ] / (log(x1))^2 Using what you mention: f(x) = y = log(base x1) of x (x1)^y = (x1)^(log base x1 of x) yielding (x1)^y = x Now taking the log of both sides: y log (x1) = log (x) y = log (x) / log (x1) 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. 
If y=u/v then yv = u
Differentiate both equations and solve for y'. 
BTW what is x "independent" of?

All times are UTC. The time now is 04:51. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2022, Jelsoft Enterprises Ltd.