Question 986461
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The only thing I can think to do is to use the base change formula and make both of the log functions in the exponent be natural logs.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(e)\ =\ \frac{\ln(e)}{\ln(3)}\ =\ \frac{1}{\ln(3)}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^{5log_3(e)ln(x)}\ =\ 3^{\frac{5\ln(x)}{\ln(3)}}\ \approx\ 3^{4.5512\cdot \ln(x)}]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \