Question 339875
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Use logarithms.  Any base will do in the final analysis, but the arithmetic will be neater in this case if you use base 2.


First, I will make the assumption that you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{\left(x\,+\,4\right)}\ =\ 4\left(16^x\right)]


as opposed to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^x\ +\ 4\ =\ 4\left(16^x\right)]


First note that *[tex \Large 16\ =\ 2^4]


Hence your equation becomes:


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


Now take the base 2 logarithm of both sides:


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


Use the following laws of logarithms to simplify the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


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


Next note that *[tex \Large \log_b(b)\ =\ 1\ \forall\ b\ \in\ \mathbb{R},\ b\ >\ 0]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 4\ =\ \log_2\left(4\right)\ +\ 4x]


But also recall that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]


So we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(4)\ =\ 2]


and then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 4\ =\ 2\ +\ 4x]


Which you can solve by ordinary means.


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