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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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    $\begingroup$ There are formulas for solving cubic equations, but there does not seem to be any "nice" solution in this case. $\endgroup$ – Per Manne Oct 17 '12 at 16:52
  • $\begingroup$ Yes, very good! Can you check the exercise again? Wasn't there a $2$ or $\sqrt x$ somewhere? Wolfram Alpha says, has one real solution but ugly: $x=0.893289..$ $\endgroup$ – Berci Oct 17 '12 at 16:53
  • $\begingroup$ @Berci Thanks, I copied it correctly. It looks like they are looking for the ugly solution! $\endgroup$ – Justin Brown Oct 17 '12 at 17:11
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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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  • $\begingroup$ Okay, I put the equation in Wolfram Alpha and got the same root, thank you. Do you have a link that explains how to get a numeric solution? $\endgroup$ – Justin Brown Oct 17 '12 at 17:14
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    $\begingroup$ @JustinBrown: There are many methods, discussed in numerical analysis texts. The simplest to explain is bisection. We know that $f(0) \lt 0$ and $f(1) \gt 0$, so there is a root in there. Check $f(0.5)=-2.625$ and we know the root is in $(0.5,1)$, then check $f(0.75)$ and so on. Keep going until the interval is small enough. There are other methods that converge faster. $\endgroup$ – Ross Millikan Oct 17 '12 at 17:22
  • $\begingroup$ I'm familiar with that technique I just didn't realize that is what you meant. Thank you! $\endgroup$ – Justin Brown Oct 17 '12 at 17:31
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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 $$\sqrt[3]{\frac{3}{2} - \sqrt{\frac{211}{108}}} + \sqrt[3]{\frac{3}{2} + \sqrt{\frac{211}{108}}} -1$$

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