Question 382247
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9^{2x}\ =\ 1667]


Take the log of both sides, any base you like.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(9^{2x}\right)\ =\ \ln(1667)]


Use


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ln\left(9\right)\ =\ \ln(1667)]


Solve by ordinary means:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln(1667)}{2\ln(9)}]


The rest is just calculator work.


Showing that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ln(1667)}{2\ln(9)}\ =\ \frac{\log_{10}(1667)}{2\log_{10}(9)}]


to get an intuitive sense that my claim about the choice of log base being immaterial to the outcome is left as an exercise for the student.


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