Question 434696
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^{x\,-\,1}\ =\ 3^{2x}]


First note: *[tex \Large 3^{2x}\ =\ 9^x] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^{x\,-\,1}\ =\ 9^x]


Take the log of both sides (any base, it doesn't matter)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(4^{x\,-\,1}\right)\ =\ \ln\left(9^x\right)]


Use the laws of logs:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(x\ -\ 1)\ln\left(4\right)\ =\ x\ln\left(9\right)]


Distribute and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ln(4)\ -\ x\ln(9)\ =\ \ln(4)]


Factor out the x


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(\ln(4)\ -\ \ln(9)\right)\ =\ \ln(4)]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(\ln\left(\frac{4}{9}\right)\right)\ =\ \ln(4)]


Solve for x:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln(4)}{\ln\left(\frac{4}{9}\right)}]


Which is the exact simplifed answer, or you can use your calculator to get a numeric approximation to whatever precision is appropriate to your application.


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