# How do you solve Log x + log(x-9)=1?

Aug 10, 2015

$\textcolor{red}{x = 10}$

#### Explanation:

 logx+log(x−9)=1

Recall that $\log a + \log b = \log \left(a b\right)$, so

 logx+log(x−9)=log(x(x-9))=log(x^2-9x)

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

Convert the logarithmic equation to an exponential equation.

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

Remember that ${10}^{\log} x = x$, so

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

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

$\left(x - 10\right) \left(x + 1\right) = 0$

$x - 10 = 0$ and $x + 1 = 0$

$x = 10$ and $x = - 1$

Check:

 logx+log(x−9)=1

If $x = 10$,

 log10+log(10−9)=1

$\log 10 + \log 1 = 1$

$1 + 0 = 1$

$1 = 1$

$x = 10$ is a solution.

If $x = - 1$,

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

This is impossible, because the logarithm of a negative number is undefined.