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?
Best Answer
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 $1.$