# How do you find the inverse of y = 1 + log_10 (x + 2) ?

Oct 10, 2015

${f}^{-} 1 \left(x\right) = \frac{{10}^{x} - 20}{10}$

#### Explanation:

$1$ can be written as ${\log}_{10} 10$.

$y = {\log}_{10} 10 + {\log}_{10} \left(x + 2\right)$

Using the Product Law which states that ${\log}_{b} A C = {\log}_{b} A + {\log}_{b} C$,

$y = {\log}_{10} \left(10 \cdot \left[x + 2\right]\right)$
$y = {\log}_{10} 10 x + 20$

Since ${\log}_{a} b = c$ can be written as $b = {a}^{c}$,

$10 x + 20 = {10}^{y}$

Make $x$ the subject.
$10 x = {10}^{y} - 20$
$x = \frac{{10}^{y} - 20}{10}$

Change $y$ to $x$ in your final answer.

${f}^{-} 1 \left(x\right) = \frac{{10}^{x} - 20}{10}$