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

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.

Answer 2

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 $\mathbb{R}^+$.

Answer 3

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 $x\neq{-1}$

$\therefore x=10$

Answer 4

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 ($x_-<9$!).