Question 518479
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^{x\,+\,2}\ =\ 3^x\ +\ 3^2]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^n\,\cdot\,a^m\ =\ a^{n\ +\ m}]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^x\,\cdot\,3^2\ =\ 3^x\ +\ 3^2]


Add *[tex \Large -3^x] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^x\,\cdot\,3^2\ -\ 3^x\ =\ 3^2]


Factor out *[tex \Large 3^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^x\left(3^2\ -\ 1\right)\ =\ 3^2]


which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^x\left(8\right)\ =\ 9]


and then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^x\ =\ \frac{9}{8}]


take the base 3 logarithm of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(3^x\right)\ =\ \log_3\left(\frac{9}{8}\right)]


Then use *[tex \Large \log_b(x^n)\ =\ n\log_b(x)] and *[tex \Large \log_b(b)\ =\ 1] to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \log_3\left(\frac{9}{8}\right)]


which should be just fine for an exact answer.  If you want a numeric approximation you can enter =LOG(9/8,3) into an Excel (Windows) or Numbers (Mac) spreadsheet cell and hit Enter, or you can use a regular scientific calculator.  If you use a calculator, you will almost certainly have to change base because most calculators only have base 10 and natural (base e) logs.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left(\frac{9}{8}\right)\ =\ \frac{\log\left(\frac{9}{8}\right)}{\log(3)}\ =\ \frac{\ln\left(\frac{9}{8}\right)}{\ln(3)}]


On the other hand, if you have to look up your logs in a table you will want to take this additional step based on the fact that the log of the quotient is the difference of the logs and the basic definition of logs which says that base raised to the log function value equals the argument:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ x\ =\ \log_3(9)\ -\ \log_3(8)\ =\ 2\ -\ \log_3(8)\ =\ 2\ -\ \frac{\ln(8)}{\ln(3)}]
 

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