Question 804377
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I presume the "+ 8" is in the exponent.  Please use parentheses in the future so that I don't have to guess what you mean.  Parentheses would have obviated the need to spell out "times" also.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2^{2x}\right)\left(4^{4x\,+\,8}\right)\ =\ 64]


Note that *[tex \Large (b^n)^m\ =\ b^{nm}], so *[tex \Large 2^{2x}\ =\ 4^x]


And *[tex \Large b^nb^m\ =\ b^{n+m}], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4^x)(4^{4x+8})\ =\ 4^{5x\,+\,8}\ =\ 64]


Take the base 4 log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4\left(4^{5x\,+\,8}\right)\ =\ \log_4(64)]


But *[tex \Large \log_b(x^n)\ =\ n\log_b(x)], and since *[tex \Large 64\ =\ 4^3] we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (5x\ +\ 8)\log_4(4)\ =\ 3]


Then since *[tex \Large \log_b(b)\ =\ 1]  (See why I picked base 4 for the log?)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ +\ 8\ =\ 3]


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


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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