Question 764142
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(2^{5x-1}\right)\ =\ \log_2(9)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (5x\ -\ 1)\log_2(2)\ =\ \log_2(9)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ -\ 1\ =\ \log_2(9)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ =\ \log_2(9)\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\log_2(9)\ +\ 1}{5}]


Which is just fine if all you need is the exact answer.  However, if you need to calculate a numeric approximation AND you don't have a calculator that will do base 2 logs, you can always start by taking either the base 10 log or the natural log.  Using *[tex \LARGE \ln] results in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln(9)\ +\ \ln(2)}{5\ln(2)}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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