Question 320816
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^x\ =\ 7\,\cdot\,3^x]


Take the natural log of both sides (actually, it can be the log to any base as it won't matter in the end).


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(5^x\right)\ =\ \ln\left(7\,\cdot\,3^x\right)]


The log of the product is the sum of the logs:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(5^x\right)\ =\ \ln\left(7\right)\ +\ \ln\left(3^x\right)]


Add *[tex \LARGE -\ln\left(3^x\right)] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(5^x\right)\ -\ \ln\left(3^x\right)\ =\ \ln\left(7\right)]


Then use:


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


so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ln\left(5\right)\ -\ x\ln\left(3\right)\ =\ \ln\left(7\right)]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(\ln\left(5\right)\ -\ \ln\left(3\right)\right)\ =\ \ln\left(7\right)]


The difference of the logs is the log of the quotient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ln\left(\frac{5}{3}\right)\ =\ \ln\left(7\right)]


Finally, multiply by *[tex \LARGE \frac{1}{\ln\left(\frac{5}{3}\right)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln\left(7\right)}{\ln\left(\frac{5}{3}\right)}]


Use the calculator for a numerical approximation if necessary.


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