Question 730140
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I have to presume you mean *[tex \LARGE 3^{\small x}\LARGE\ =\ \sqrt{5^{\small x\,-\,2}}] because the alternate interpretation, namely *[tex \LARGE 3^{\small x}\LARGE\ =\ \sqrt{5^{\small x}\LARGE\,-\,2}], has no real number solutions.


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3^{\small 2x}\LARGE\ =\ 5^{\small x\,-\,2}]


Take the log of both sides, any base, it really doesn't matter.  I'll use the natural log.


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


Use *[tex \LARGE \log_b(x^n)\ =\ n\log_b(x)] to write:


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


Then keeping in mind that *[tex \LARGE \ln(3)] and *[tex \LARGE \ln(5)] are nothing more than irrational numerical constants,  we play a little algebra music:


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


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


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


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


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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