Question 627046
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Note:  *[tex \LARGE 3^{2x}\ =\ 9^x]


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


Take the natural log of both sides (actually, any base will do as long as you use the same base on both sides of the equation)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(2^{x\,-\,2}\right)\ =\ \ln\left(9^x\right)]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 2)\ln\left(2\right)\ =\ x\ln\left(9\right)]


Multiply both sides by the reciprocal of *[tex \LARGE \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 2)\ =\ x\frac{\ln\left(9\right)}{\ln\left(2\right)}]


*[tex \LARGE x] terms on the left, constants on the right, just like any other single variable equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ x\frac{\ln\left(9\right)}{\ln\left(2\right)}\ =\ 2]


Factor out *[tex \LARGE x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(1\ -\ \frac{\ln\left(9\right)}{\ln\left(2\right)}\right)\ =\ 2]


Multiply by the reciprocal of the coefficient on *[tex \LARGE x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{2}{1\ -\ \frac{\ln\left(9\right)}{\ln\left(2\right)}}]


The rest is just calculator work.


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