# What is the inverse of log (x/2) ?

Mar 28, 2016

Assuming this is base-10 logarithm, the inverse function is
$y = 2 \cdot {10}^{x}$

#### Explanation:

Function $y = g \left(x\right)$ is called inversed to function $y = f \left(x\right)$ if and only if
$g \left(f \left(x\right)\right) = x$ and $f \left(g \left(x\right)\right) = x$

Just as a refreshment on logarithms, the definition is:
${\log}_{b} \left(a\right) = c$ (for $a > 0$ and $b > 0$)
if and only if $a = {b}^{c}$.
Here $b$ is called a base of a logarithm, $a$ - its argument and $c$ - its balue.

This particular problem uses $\log \left(\right)$ without explicit specification of the base, in which case, traditionally, base-10 is implied. Otherwise the notation ${\log}_{2} \left(\right)$ would be used for base-2 logarithms and $\ln \left(\right)$ would be used for base-$e$ (natural) logarithms.

When $f \left(x\right) = \log \left(\frac{x}{2}\right)$ and $g \left(x\right) = 2 \cdot {10}^{x}$ we have:
$g \left(f \left(x\right)\right) = 2 \cdot {10}^{\log \left(\frac{x}{2}\right)} = 2 \cdot \frac{x}{2} = x$
$f \left(g \left(x\right)\right) = \log \left(\frac{2 \cdot {10}^{x}}{2}\right) = \log \left({10}^{x}\right) = x$