Find the Inverse 16^x

$16^{x}$

Step 1

Interchange the variables.

$x = 16^{y}$

Step 2

Solve for $y$ .

Rewrite the equation as $16^{y} = x$ .

$16^{y} = x$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (16^{y}) = \ln (x)$

Expand $\ln (16^{y})$ by moving $y$ outside the logarithm.

$y \ln (16) = \ln (x)$

Divide each term in $y \ln (16) = \ln (x)$ by $\ln (16)$ and simplify.

Divide each term in $y \ln (16) = \ln (x)$ by $\ln (16)$ .

$\frac{y \ln (16)}{\ln (16)} = \frac{\ln (x)}{\ln (16)}$

Simplify the left side.

Cancel the common factor of $\ln (16)$ .

Cancel the common factor.

$\frac{y \ln (16)}{\ln (16)} = \frac{\ln (x)}{\ln (16)}$

Divide $y$ by $1$ .

$y = \frac{\ln (x)}{\ln (16)}$

Step 3

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{\ln (x)}{\ln (16)}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (x)}{\ln (16)}$ is the inverse of $f (x) = 16^{x}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} \cdot 16^{x}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot 16^{x} = \frac{\ln (16^{x})}{\ln (16)}$

Expand $\ln (16^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} \cdot 16^{x} = \frac{x \ln (16)}{\ln (16)}$

Cancel the common factor of $\ln (16)$ .

Cancel the common factor.

$f^{- 1} \cdot 16^{x} = \frac{x \ln (16)}{\ln (16)}$

Divide $x$ by $1$ .

$f^{- 1} \cdot 16^{x} = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{\ln (x)}{\ln (16)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x)}{\ln (16)}) = 16^{\frac{\ln (x)}{\ln (16)}}$

Use the change of base rule $\frac{\log_{b} (x)}{\log_{b} (a)} = \log_{a} (x)$ .

$f (\frac{\ln (x)}{\ln (16)}) = 16^{\log_{16} (x)}$

Exponentiation and log are inverse functions.

$f (\frac{\ln (x)}{\ln (16)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x)}{\ln (16)}$ is the inverse of $f (x) = 16^{x}$ .

$f^{- 1} (x) = \frac{\ln (x)}{\ln (16)}$

Find the Inverse 16^x

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