Question 248322
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I suppose you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16^{2x-5}\ =\ 8^{x-3}]


For future reference, use ^ for exponentiation and use parentheses to make certain there is no ambiguity.  For example, what you wrote could have been interpreted:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16^{2}x-5\ =\ 8^{x}-3]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16^{2x}-5\ =\ 8^{x-3}]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16^{2x-5}\ =\ 8^{x}-3]


Back to my original presumption:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16^{2x-5}\ =\ 8^{x-3}]


Note that *[tex \LARGE 16\ =\ 2^4] and *[tex \LARGE 8\ =\ 2^3] and remember that *[tex \LARGE (a^m)^n\ = a^{mn}]


So write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^4^{2x-5}\ =\ 2^3^{x-3}]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{8x-20}\ =\ 2^{3x-9}]


Since


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


We can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x\ -\ 20\ =\ 3x\ -\ 9]


Solve for *[tex \Large x]



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