Originally Posted by

**Prove It** An alternate method:

First note that $\displaystyle x > 0$ in order to not have a negative in either logarithm...

$\displaystyle \log_9(x + 1) = \frac{1}{2} + \log_9{(x)}$

$\displaystyle \log_9(x + 1) - \log_9(x) = \frac{1}{2}$

$\displaystyle \log_9{\left(\frac{x + 1}{x}\right)} = \frac{1}{2}$

$\displaystyle \frac{x + 1}{x} = 9^{\frac{1}{2}}$

$\displaystyle \frac{x + 1}{x} = \{-3, 3\}$

$\displaystyle 1 + \frac{1}{x} = \{-3, 3\}$

$\displaystyle \frac{1}{x} =\{-4, 2\}$

$\displaystyle x = \left\{-\frac{1}{4}, \frac{1}{2}\right\}$.

Now since $\displaystyle x > 0$, that means the only solution is $\displaystyle x = \frac{1}{2}$.