Question 254226
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^{x\,-\,4}\ =\ 7^{-7x}]


Take the base-10 log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(6^{x\,-\,4}\right)\ =\ \log\left(7^{-7x}\right)]


Apply


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\,-\,4)\log\left(6\right)\ =\ -7x\cdot\log\left(7\right)]


Multiply both sides by *[tex \Large \frac{1}{\log(6)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\,-\,4\ =\ \left(\frac{-7\cdot\log\left(7\right)}{\log(6)}\right)x]


Add *[tex \Large 4\ +\ \left(\frac{7\cdot\log\left(7\right)}{\log(6)}\right)x] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ + \left(\frac{7\cdot\log\left(7\right)}{\log(6)}\right)x\ =\ 4]


Factor the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(1\ + \frac{7\cdot\log\left(7\right)}{\log(6)}\right)\ =\ 4]


Multiply both sides by the reciprocal of the constant coefficient in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{4\cdot\log(6)}{\log(6)\ +\ 7\cdot\log(7)}\ \approx\ 0.464996]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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