Question 461708
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Take the log of both sides -- doesn't matter what base:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \5^{x\,-\,1}\ =\ 2\left(3^x\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ln\left(\5^{x\,-\,1}\right)\ =\ \ln\left(2\left(3^x\right)\right)]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ln\left(\5^{x\,-\,1}\right)\ =\ \ln2\ +\ \ln\left(3^x\right)]


Use


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ (x\ -\ 1)\ln\left(\5\right)\ =\ \ln2\ +\ x\ln\left(3\right)]


Distribute and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ x\ln(5)\ -\ x\ln(3)\ =\ \ln(2)\ +\ \ln(5)]


and finally, using "the sum of the logs is the log of the product"


*[tex \LARGE \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln(10)}{\ln(5)\ -\ \ln(3)}]


You can also use "the difference of the logs is the log of the quotient" to further simplify(?):


*[tex \LARGE \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln(10)}{\ln\left(\frac{5}{3}\right)}]


Either way should be acceptable, but that will depend on your instructor.


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