Question 1037819
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Rules needed:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{m+n}\ =\ a^m\,\cdot\,a^n]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{mn}\ =\ a^m^n]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{-n}\ =\ \frac{1}{a^n}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x^n)\ =\ n\cdot\log(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ b\ \Rightarrow\ \log_c(a)\ =\ \log_c(b)]


The sum of the logs is the log of the product.  The difference of the logs is the log of the quotient.


Definition:  *[tex \LARGE \log(x)\ =\ \log_{10}(x)]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{x\,-\,6}\ =\ 7^{2x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(2^{x\,-\,6}\right)\ =\ \log\left(7^{2x}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\,-\,6)\log\left(2\right)\ =\ x\log\left(7^2\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\log\left(2\right)\ -\ 6\log\left(2\right)\ =\ x\log\left(49\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\log\left(2\right)\ -\ x\log\left(49\right)\ =\ 6\log\left(2\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(\log\left(2\right)\ -\ \log\left(49\right)\right)\ =\ \log\left(2^6\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\log\left(64\right)}{\log\left(2\right)\ -\ \log\left(49\right)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\log\left(64\right)}{\log\left(\frac{2}{49}\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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