# How do you find the inverse of  f(x)=log(x+15)?

Feb 6, 2016

Set $y = f \left(x\right)$ and solve with respect to $x$ hence you get

$y = \log \left(x + 15\right) \implies x + 15 = {10}^{y} \implies x = {10}^{y} - 15$

Now it is ${f}^{-} 1 \left(x\right) = {10}^{x} - 15$

Assuming that $\log 10 = 1$

Feb 7, 2016

${f}^{-} 1 \left(x\right) = {10}^{x} - 15$

#### Explanation:

For any inverse function of $f \left(x\right) , {f}^{-} 1 \left(x\right)$
$f \left({f}^{-} 1 \left(x\right)\right) = x$
$f \left({f}^{-} 1 \left(x + 15\right)\right) = \log \left({f}^{-} 1 \left(x + 15\right)\right) = \log \left({f}^{-} 1 \left(x\right) + 15\right)$
${10}^{x} = {f}^{-} 1 \left(x\right) + 15$
${f}^{-} 1 \left(x\right) = {10}^{x} - 15$

Alternatively:
u = x + 15; x = u -15
y = log_b u; b^y = u; b = 10; 10^y = u  substitute for u
10^y = x+15; x = 10^y -15 replace y by x and write
${f}^{-} 1 \left(x\right) = {10}^{x} - 15$ which is exactly as above