[Math] Using logarithmic identities to solve $\log x – 1 = -\log (x-9)$

Question

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

Answer 1

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 $x\neq{-1}$

$\therefore x=10$