Question 518402
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{2x}\ =\ 7e^{x\ -\ 12}]

Take the natural log of both sides:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{2x}\right)\ =\ \ln\left(7e^{x\ -\ 12}\right)]

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

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{2x}\right)\ =\ \ln\left(7\right)\ +\ \ln\left(e^{x\ -\ 12}\right)]

Use

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

and

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(e)\ =\ 1]

to write

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ \ln(7)\ +\ x\ -\ 12]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \ln(7)\ -\ 12]

Is the exact answer.  A numerical approximation is right there in your calculator.

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