Question 681248
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The difference of the logs is the log of the product.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x)\ -\ \log_2(x\,+\,4)\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(\frac{x}{x\,+\,4}\right)\ =\ 3]


Use the definition of the logarithm function:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3 = \log_2\left(\frac{x}{x\,+\,4}\right) \ \ \Rightarrow\ \ 2^3 = \frac{x}{x\,+\,4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8\ =\ \frac{x}{x\,+\,4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{32}{7}]


But the domain of the log function is restricted to positive reals.  Therefore the solution set is empty.


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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