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?
On
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 $$\sqrt[3]{\frac{3}{2} - \sqrt{\frac{211}{108}}} + \sqrt[3]{\frac{3}{2} + \sqrt{\frac{211}{108}}} -1$$
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 $1.$