# How do you solve  log2 x = log (x + 16) - 1?

Mar 16, 2016

$x = \frac{16}{19}$

#### Explanation:

$1$. Start by bringing $\log \left(x + 16\right)$ to the left side of the equation.

$\log \left(2 x\right) = \log \left(x + 16\right) - 1$

$\log \left(2 x\right) - \log \left(x + 16\right) = - 1$

$2$. Using the log property, ${\log}_{\textcolor{p u r p \le}{b}} \left(\frac{\textcolor{red}{m}}{\textcolor{b l u e}{n}}\right) = {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right) - {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{n}\right)$, simplify the equation.

$\log \left(\frac{2 x}{x + 16}\right) = - 1$

$3$. Convert to exponential form.

${10}^{- 1} = \left(\frac{2 x}{x + 16}\right)$

$4$ Solve for $x$.

$\frac{1}{10} = \left(\frac{2 x}{x + 16}\right)$

$\frac{x + 16}{10} = 2 x$

$x + 16 = 20 x$

$19 x = 16$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{16}{19} \textcolor{w h i t e}{\frac{a}{a}} |}}}$