# What is the inverse of y = 3log_2 (4x)-2  ?

Nov 10, 2015

${f}^{- 1} \left(x\right) = {4}^{- \frac{2}{3}} \cdot {2}^{\frac{x}{3}}$

#### Explanation:

First, switch $y$ and $x$ in your equation:

$x = 3 {\log}_{2} \left(4 y\right) - 2$

Now, solve this equation for $y$:

$x = 3 {\log}_{2} \left(4 y\right) - 2$
$\iff x + 2 = 3 {\log}_{2} \left(4 y\right)$
$\iff \frac{x + 2}{3} = {\log}_{2} \left(4 y\right)$

The inverse function of ${\log}_{2} \left(a\right)$ is ${2}^{a}$, so apply this operation to both sides of the equation to get rid of the logarithm:

$\iff {2}^{\frac{x + 2}{3}} = {2}^{{\log}_{2} \left(4 y\right)}$
$\iff {2}^{\frac{x + 2}{3}} = 4 y$

Let's simplify the expression on the left side using the power rules ${a}^{n} \cdot {a}^{m} = {a}^{n + m}$ and ${a}^{n \cdot m} = {\left({a}^{n}\right)}^{m}$:

${2}^{\frac{x + 2}{3}} = {2}^{\frac{x}{3} + \frac{2}{3}} = {2}^{\frac{x}{3}} \cdot {2}^{\frac{2}{3}} = {2}^{\frac{x}{3}} \cdot {\left({2}^{2}\right)}^{\frac{1}{3}} = {4}^{\frac{1}{3}} \cdot {2}^{\frac{x}{3}}$

Let's get back to our equation:

${2}^{\frac{x + 2}{3}} = 4 y$
$\iff {4}^{\frac{1}{3}} \cdot {2}^{\frac{x}{3}} = 4 y$
$\iff {4}^{\frac{1}{3}} / 4 \cdot {2}^{\frac{x}{3}} = y$
$\iff {4}^{- \frac{2}{3}} \cdot {2}^{\frac{x}{3}} = y$

You are done. The only thing left to do is replacing $y$ with ${f}^{- 1} \left(x\right)$ for a more formal notation:

for
$f \left(x\right) = 3 {\log}_{2} \left(4 x\right) - 2$,
the inverse function is
${f}^{- 1} \left(x\right) = {4}^{- \frac{2}{3}} \cdot {2}^{\frac{x}{3}}$.

Hope that this helped!