Question 1053172
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Note:  I follow the convention that *[tex \Large \log] means *[tex \Large \log_{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 13^{-x\,+\,6}\ =\ 15^{-10x}]


Take the log of both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(13^{-x\,+\,6}\right)\ =\ \log\left(15^{-10x}\right)]


Use *[tex \Large \log_b\left(x^n\right)\ =\ n\log_b(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-x\,+\,6)\log\left(13\right)\ =\ (-10x)\log\left(15)\right)]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ -\ 6}{x}\ =\ \frac{10\log(15)}{\log(13)}\ =\ \frac{\log\left(15^{10}\right)}{\log(13)}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \frac{6}{x}\ =\ \frac{\log\left(15^{10}\right)}{\log(13)}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6}{x}\ =\ 1\ -\ \frac{\log\left(15^{10}\right)}{\log(13)}\ =\ \frac{\log(13)}{\log(13)}\ -\ \frac{\log\left(15^{10}\right)}{\log(13)}\ =\ \frac{\log(13)\ -\ \log\left(15^{10}\right)}{\log(13)}]


The difference of the logs is the log of the quotient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6}{x}\ = \frac{\log\left(\frac{13}{15^{10}}\right)}{\log(13)}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ = \frac{\log\left(\frac{13}{15^{10}}\right)}{6\log(13)}\ =\ \frac{\log\left(\frac{13}{15^{10}}\right)}{\log(13^6)}]


Take the reciprocal


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\log(13^6)}{\log\left(\frac{13}{15^{10}}\right)}\ =\ \frac{\log(4,826,809)}{\log\left(\frac{13}{576,650,390,625}\right)}]



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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