Question 242715
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Your clarification just confused the issue.  Now I'm not sure whether you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^{3x^{-1}}\ =\ 5^{5x^{-3}}]


Which is what you initially wrote, or if you meant


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5^{3x-1}\ =\ 5^{5x-3}]


which is what your clarification says.


Be that as it may, you solve either one the same way.  Since the base number is the same in each case, namely 5, you can use the fact that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^m\ =\ a^n\ \Leftrightarrow\ m\ =\ n]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^{-1}\ =\ 5x^{-3}]


and then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{x}\ =\ \frac{5}{x^3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\sqrt{\frac{5}{3}}\ =\ \pm\frac{\sqrt{15}}{3}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ -\ 1\ =\ 5x\ -\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1]


again, depending on the correct form of your problem.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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