# How do you solve log (x + 8) = 1 + log (x - 10)?

Dec 14, 2015

$x = 12$

#### Explanation:

Begin by moving both of the $\log$ terms to the left hand side.

$\log \left(x + 8\right) - \log \left(x - 10\right) = 1$

Now we can use the division rule for logarithms to combine both terms into one. The division rule states that;

$\log \left(\frac{m}{n}\right) = \log \left(m\right) - \log \left(n\right)$

Letting $m = x + 8$ and $n = x - 10$, we get;

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

Since we are working with a common $\log$ it is base ten. That means that the part inside of the parenthesis is equal to $10$ raised to the power of the right hand side, or;

${10}^{1} = \frac{x + 8}{x - 10}$

Now we just need to do some algebra to solve for $x$. First, multiply both sides by $\left(x - 10\right)$.

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

Now multiply the $10$ through the parenthesis.

$10 x - 100 = x + 8$

Subtract $x$ and add $100$ to both sides.

$9 x = 108$

Finally, divide both sides by $9$ to find $x$.

$x = 12$