I'm getting stuck trying to solve this logarithmic equation:
$ \log( \sqrt{4-x} ) - \log( \sqrt{x+3} ) = \log(x) $ I understand that the first and second terms can be combined & the logarithms share the same base so one-to-one properties apply and I get to: $ x = \frac{\sqrt{4-x}}{ \sqrt{x+3} } $ Now if I square both sides to remove the radicals: $ x^2 = \frac{4-x}{x+3} $ Then: $ x^2(x+3) = 4-x $ $ x^3 +3x^2 + x - 4 = 0 $
Is this correct so far? How do I solve for x from here?