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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?

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    @Berci Thanks, I copied it correctly. It looks like they are looking for the ugly solution!2012-10-17

2 Answers 2

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Fine so far. I would just use Wolfram Alpha, which shows there is a root about $0.89329$. The exact value is a real mess. I tried the rational root theorem, which failed. If I didn't have Alpha, I would go for a numeric solution. You can see there is a solution in $(0,1)$ because the left side is $-4$ at $0$ and $+1$ at $1.$

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    I'm familiar with that technique I just didn't realize that is what you meant. Thank you!2012-10-17
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It is correct so far.

There is clearly a root between $0$ and $1$. Either use numerical methods to find it is about $0.893289$ or (not recommended) solve the cubic to get $\sqrt[3]{\frac{3}{2} - \sqrt{\frac{211}{108}}} + \sqrt[3]{\frac{3}{2} + \sqrt{\frac{211}{108}}} -1$