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Solve for X (logs):
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2in(x-5) = (x-2) +in (x+3)
2ln(x-5)- ln(x+3) = x-2 ln (x-5)^2 / (x+3) = x - 2 (x-5)^2 / (x+ 3) = e^(x-2)
Then from here?
shoot - i'm stumped - where to go from here??
this type of equation can be solved by trial and improvement the answer is x = 2 which i found using wolfram alpha check it out : http://www.wolframalpha.com/input/?i=%28x-5%29%5E2+%2F+%28x%2B+3%29+++%3D++e%5E%28x-2%29
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there could be a better way to solve this but can't think of one!!!!
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