Question 799369
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You started off on the right foot, but you will have a much easier time of it if you take the base 2 log of both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8^{x\,-\,1}\ =\ 16^{x\,+\,3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{3(x\,-\,1)}\ =\ 2^{4(x\,+\,3)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(2^{3(x\,-\,1)}\right)\ =\ \log_2\left(2^{4(x\,+\,3)}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(3(x\,-\,1)\right)\log_2(2)\ =\ \left(4(x\,+\,3)\right)\log_2(2)]


But since *[tex \LARGE \log_b(b)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3(x\,-\,1)\ =\ 4(x\,+\,3)]


from which it follows:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -15]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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