Question 534468
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You really need to use parentheses so that your statements aren't ambiguous.   I'm guessing you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^{x\ -\ 10}\ \cdot\ 2^{x\ -\ 74}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2^2\right)^{x\ -\ 10}\ \cdot\ 2^{x\ -\ 74}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{2x\ -\ 20}\ \cdot\ 2^{x\ -\ 74}\ =\ 1]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{3x\ -\ 94}\ =\ 1]


Take the base 2 log of both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(2^{3x\ -\ 94}\right)\ =\ \log_2(1)]


Use the laws of logs:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3x\ -\ 94)\log_2\left(2\right)\ =\ \log_2(1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ -\ 94\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ =\ 94]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{94}{3}]


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