Question 725775
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The difference of the logs is the log of the quotient.


*[tex \LARGE \log_{\small{\frac{1}{3}}\LARGE}\left(x^2\ +\ x\right)\ -\ \log_{\small{\frac{1}{3}}\LARGE}\left(x^2\ -\ x\right)\ =\ -1]


*[tex \LARGE \log_{\small{\frac{1}{3}}\LARGE}\left(\frac{x^2\ +\ x}{x^2\ -\ x}\right)\ =\ -1]


Use the definition of the log function:


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


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1}{3}\right)^{-1}\ =\ \left(\frac{x^2\ +\ x}{x^2\ -\ x}\right)]


Then solve for *[tex \LARGE x] but note that the root of the rational expression is NOT in the domain either of one of the original log functions nor in the domain of the resulting rational function.  The solution set is the empty set.


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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