Question 261034
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^{2x\,-\,1}\ =\ 73]


Take the natural log (or base 10 log if you  like -- doesn't matter) of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(6^{2x\,-\,1}\right)\ =\ \ln\left(73\right)]


Use


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2x\ -\ 1\right)\ln\left(6\right)\ =\ \ln\left(73\right)]


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



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2x\ -\ 1\right)\ =\ \frac{\ln\left(73\right)}{\ln\left(6\right)}]


Add 1 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ \frac{\ln\left(73\right)}{\ln\left(6\right)}\ +\ 1]


Multiply both sides by *[tex \Large \frac{1}{2}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln\left(73\right)}{2\ln\left(6\right)}\ +\ \frac{1}{2}]


Use


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln\left(73\right)}{\ln\left(36\right)}\ +\ \frac{1}{2}]


A little calculator work and a round-off should get you the answer you need.


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