Question 184185
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \log_4(x) - \log_4(x+3)= -1]


The difference of the logs is the log of the quotient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \log_4\left(\frac{x}{x+3}\right)=-1]


Now use:


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


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \frac{x}{x+3}=4^{-1}=\frac{1}{4}]


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 4x = x + 3\ \ \Rightarrow\ \ 3x = 3 \ \ \Rightarrow\ \ x = 1]


John
*[tex \Large e^{i\pi} + 1 = 0]
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