Question 596808
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There are two ways to interpret the problem as you rendered it.


You either meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,\cdot\,\log_2(x\ +\ 1)\ =\ 7]


the solution to which would make a fiend from the seventh circle of Hell look attractive, or you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,\cdot\,\log_2(x)\ +\ 1\ =\ 7]


the solution to which is nice, neat, and only involves integers.  I'm going for the simple way since, at this point, I have 51% of the vote.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,\cdot\,\log_2(x)\ +\ 1\ =\ 7]


Add -1 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\,\cdot\,\log_2(x)\ =\ 6]


Multiply both sides by *[tex \Large \frac{1}{3}]


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


Use the definition of the logarithm function


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x)\ =\ 2 \ \ \Rightarrow\ \ 2^2 = x]


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


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