 mersenneforum.org Exponentiation w/ independent variable
 Register FAQ Search Today's Posts Mark Forums Read 2010-07-08, 16:55 #1 Unregistered   22×5×132 Posts 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.  2010-07-11, 18:47 #2 Primeinator   "Kyle" Feb 2005 Somewhere near M52.. 3·5·61 Posts 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?   2010-07-11, 22:27 #3 Primeinator   "Kyle" 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   2010-07-13, 08:07 #4 davieddy   "Lucan" Dec 2006 England 145128 Posts If y=u/v then yv = u Differentiate both equations and solve for y'.   2010-08-04, 06:38 #5 davieddy   "Lucan" Dec 2006 England 647410 Posts BTW what is x "independent" of?  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post siegert81 Factoring 1 2018-04-16 22:53 carpetpool Information & Answers 2 2017-11-03 07:24 ramshanker Math 2 2015-10-31 15:28 Mini-Geek Software 7 2008-01-14 17:33 Greenbank Math 5 2005-09-30 10:20

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