# How do you solve Log(x-9) = 3-Log(100x) ?

Jul 3, 2018

$\textcolor{b l u e}{x = 10}$

#### Explanation:

$\log \left(x - 9\right) = 3 - \log \left(100 x\right)$

By the laws of logarithms:

$\log \left(a b\right) = \log \left(a\right) + \log \left(b\right) \textcolor{w h i t e}{888} \left[1\right]$

$\log \left(100 x\right) = \log \left(100\right) + \log \left(x\right)$

Assuming these are base 10 logarithms:

$\log \left(100\right) + \log \left(x\right) = 2 + \log \left(x\right)$

We now have:

$\log \left(x - 9\right) = 3 - 2 - \log \left(x\right)$

$\log \left(x - 9\right) = 1 - \log \left(x\right)$

Using $\left[1\right]$

$\log \left(x - 9\right) + \log \left(x\right) = 1$

$\log \left(x \left(x - 9\right)\right) = 1$

$\log \left({x}^{2} - 9 x\right) = 1$

${10}^{\log \left({x}^{2} - 9 x\right)} = {10}^{1}$

${x}^{2} - 9 x = 10$

${x}^{2} - 9 x - 10 = 0$

Factor:

$\left(x + 1\right) \left(x - 10\right) = 0 \implies x = - 1 \mathmr{and} x = 10$

Checking solutions.

$x = - 1$

$\log \left(\left(- 1\right) - 9\right) = 3 - \log \left(100 \left(- 1\right)\right)$

$\log \left(- 10\right) = 3 - \log \left(- 100\right)$

Logarithms are only defined for real numbers if for:

$\log \left(x\right)$

$x > 0$

Therefore $- 1$ is not a solution.

For $x = 10$

$\log \left(10 - 9\right) = 3 - \log \left(100 \left(10\right)\right)$

$\log \left(1\right) = 3 - \log \left(1000\right)$

$0 = 3 - 3$

$0 = 0$

So $x = 10$ is the only solution.