Using logarithmic identities to solve $\log x - 1 = -\log (x-9)$ Could someone take a look at the following and give me a breakdown for this log equation? I'm stuck...
Solve for $x$:
$$\log x - 1 = -\log (x-9)$$
Many thanks
 A: Assuming $\log$ means base $10$ (and you can modify this to other bases):
$\log(x)-1=-\log(x-9)$
$\Longrightarrow \log(x)+\log(x-9)-1=0$
$\Longrightarrow \log\big(x\hspace{1 mm}(x-9)\big)=1$
$\Longrightarrow x\hspace{1 mm}(x-9)=10^{1}$
$\Longrightarrow x^{2}-9x-10=0$
$\Longrightarrow x=\dfrac{-(-9)\pm\sqrt{(-9)^{2}-4(1)(-10)}}{2(1)}$
$\Longrightarrow x = \dfrac{9\pm{11}}{2}$
$\Longrightarrow x=-1,10$
But $x\neq{-1}$
$\therefore x=10$
A: Presumably, you are looking for real solutions, so it is safe to assume that $x>9$. Therefore, I leave you with the hint below...
Hint: $\log a+\log b = \log(ab)$ whenever $a$ and $b$ are both positive.
A: Hint: assuming base 10 logarithms, write it as $$\log\left(\frac{x}{10}\right)=\log\left(\frac{1}{x-9}\right)$$
Then remember that $\log$ is injective on $\mathbb{R}^+$.
A: We must require $x>0$ and $x-9>0\longrightarrow x>9$, so $x>9$ as final condition. 
If $\log\equiv\log_{e}\equiv\ln$, then:
$\ln x - 1 = -\ln (x-9)$
$\ln x - \ln e= -\ln (x-9)$
$\ln {x\over e}= \ln (x-9)^{-1}$
${x\over e}={1\over x-9}$
$x^2-9x-e=0$
Solving this equation you get $x_\pm={9\pm\sqrt{81+4e}\over2}$ and $x_+$ is the solution ($x_-<9$!).
