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Help me Solve another Logarithmic Equation Please
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\(\large\color{black}{ \log_{_{\LARGE1/5}}~(x^2+x)-\log_{_{\LARGE1/5}}~(x^2-x) =-1 }\)
this is what I would do (first thing that comes up) use the rule, \(\large\color{black}{ \log_{_{\LARGE a}}~(A)-\log_{_{\LARGE a}}~(B) =\log_{_{\LARGE a}}~(\frac{A}{B}) }\)
then when you write it as a single logarithm, re-write the -1 in terms of a log base 1/5.
\[\log_{1/5} = x^2+x/x^2-x (-1) ?\]
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at first, when we write this as a single logarithm, using the aforementioned rule, |dw:1417564900448:dw|
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