Question 1166261
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(2x)\ -\ \log_2(x\,+\,1)\ =\ -3]


The difference of the logs is the log of the quotient


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\(\frac{2x}{x\,+\,1}\)\ =\ -3]


The definition of the logarithm function is:


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


So


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\ =\ \frac{2x}{x\,+\,1}]


Solve for *[tex \Large x]

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