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find the inverse of f(x)= (3^x)/(2-3^x)
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the x powers are the part that is confusing me, I know I have to solve for x and then switch the x and y... but I cant get to that poing
first..change f(x) into y \[y = \frac{3^x}{2 - 3^x}\] cross multiply \[\implies y(2 - 3^x) = 3^x\] distribute \[\implies 2y - y3^x = 3^x\] put the x terms on one side \[\implies 2y = 3^x + y3^x\] factor out \[2 = 3^x(1 + y)\] divide both sides by 1 + y \[\frac{2}{1+y} = 3^x\] change to log form \[\log_3 \left (\frac{2}{1+y}\right) = x\] does that help?
@cmgeorgia you there?
yes, I was reviewing the problem. Thank you so much! One thing to clarify, shouldn't it be 2y/ 1+y)?
ahh yes..good observation :D
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okay just making sure! Thanks again!
welcome
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