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log (5+x)-log (x-3)=log 3
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loga-logb =log(a/b) next step cancel logs and slove for x
Do not EVER "cancel logs". That doesn't mean anything. After using the logarithm property, \(\log(5+x) - \log(x-3) = \log\left(\dfrac{5+x}{x-3}\right)\)... Note that if \(\log(a) = \log(b)\;then\;a = b\), since the logarithm is monotonic increasing. This leads to... \(\dfrac{5+x}{x-3} = 3\) Thinking about Domain issues. In the original problem statement, we have... \(5+x > 0 \longrightarrow x > -5\) \(x-3 > 0 \longrightarrow x > 3\) Remembering that, whatever else happens, do NOT accept a result of 3 or less. It is wrong.
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