Question 1050558
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Let


Equation 1:  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^\varphi\ =\ 9^{x\,-\,1}]


Take the natural log of both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left( e^\varphi\right)\ =\ \ln\left(9^{x\,-\,1}\right)]


Use *[tex \Large \log_b(a^n)\ =\ n\log_b(a)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi\ln\left( e\right)\ =\ \ln\left(9^{x\,-\,1}\right)]


Use *[tex \Large \log_b(b)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi\ =\ \ln\left(9^{x\,-\,1}\right)]


Now just substitute for *[tex \Large \varphi] in Equation 1.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9^{x\,-\,1}\ =\ e^{\ln\left(9^{x\,-\,1}\right)}]


or you could also write it as


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9^{x\,-\,1}\ =\ e^{\left(x\,-\,1\right)\ln\left(9^\right)}]


depending on which makes any subsequent calculations simpler.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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