# How do you solve  log x- log(x-8)=1?

$x = \frac{80}{9} = 8 \frac{8}{9}$

#### Explanation:

The given:$\log x - \log \left(x - 8\right) = 1$

This means the common logarithm of base 10:

${\log}_{10} x - {\log}_{10} \left(x - 8\right) = 1$

Also ${\log}_{10} 10 = 1$

therefore

${\log}_{10} x - {\log}_{10} \left(x - 8\right) = {\log}_{10} 10$

${\log}_{10} \left(\frac{x}{x - 8}\right) = {\log}_{10} 10$

Take Antilogarithm of both sides of the equation

$A n t i \log \left({\log}_{10} \left(\frac{x}{x - 8}\right)\right) = A n t i \log \left({\log}_{10} 10\right)$

means

${10}^{{\log}_{10} \left(\frac{x}{x - 8}\right)} = {10}^{{\log}_{10} 10}$

means

$\frac{x}{x - 8} = 10$

solve for $x$ now

$x = 10 \left(x - 8\right)$

$x = 10 x - 80$

$x - 10 x = - 80$

$- 9 x = - 80$

$\frac{- 9 x}{-} 9 = \frac{- 80}{-} 9$

$x = \frac{80}{9} = 8 \frac{8}{9}$

Have a nice day !!! from the Philippines..