Find the Inverse y=5^(x/2)

$y = 5^{\frac{x}{2}}$

Interchange the variables.

$x = 5^{\frac{y}{2}}$

Solve for $y$ .

Rewrite the equation as $5^{\frac{y}{2}} = x$ .

$5^{\frac{y}{2}} = x$

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

$\ln (5^{\frac{y}{2}}) = \ln (x)$

Expand the left side.

Expand $\ln (5^{\frac{y}{2}})$ by moving $\frac{y}{2}$ outside the logarithm.

$\frac{y}{2} \ln (5) = \ln (x)$

Combine $\frac{y}{2}$ and $\ln (5)$ .

$\frac{y \ln (5)}{2} = \ln (x)$

Multiply both sides of the equation by $\frac{2}{\ln (5)}$ .

$\frac{2}{\ln (5)} \cdot \frac{y \ln (5)}{2} = \frac{2}{\ln (5)} \ln (x)$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{2}{\ln (5)} \cdot \frac{y \ln (5)}{2}$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2}{\ln (5)} \cdot \frac{y \ln (5)}{2} = \frac{2}{\ln (5)} \ln (x)$

Rewrite the expression.

$\frac{1}{\ln (5)} (y \ln (5)) = \frac{2}{\ln (5)} \ln (x)$

Cancel the common factor of $\ln (5)$ .

Factor $\ln (5)$ out of $y \ln (5)$ .

$\frac{1}{\ln (5)} (\ln (5) y) = \frac{2}{\ln (5)} \ln (x)$

Cancel the common factor.

$\frac{1}{\ln (5)} (\ln (5) y) = \frac{2}{\ln (5)} \ln (x)$

Rewrite the expression.

$y = \frac{2}{\ln (5)} \ln (x)$

Simplify the right side.

Combine $\frac{2}{\ln (5)}$ and $\ln (x)$ .

$y = \frac{2 \ln (x)}{\ln (5)}$

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

$f^{- 1} (x) = \frac{2 \ln (x)}{\ln (5)}$

Verify if $f^{- 1} (x) = \frac{2 \ln (x)}{\ln (5)}$ is the inverse of $f (x) = 5^{\frac{x}{2}}$ .

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 $f^{- 1} \cdot 5^{\frac{x}{2}}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot 5^{\frac{x}{2}} = \frac{2 \ln (5^{\frac{x}{2}})}{\ln (5)}$

Expand $\ln (5^{\frac{x}{2}})$ by moving $\frac{x}{2}$ outside the logarithm.

$f^{- 1} \cdot 5^{\frac{x}{2}} = \frac{2 (\frac{x}{2} \cdot \ln (5))}{\ln (5)}$

Cancel the common factor of $\ln (5)$ .

Cancel the common factor.

$f^{- 1} \cdot 5^{\frac{x}{2}} = \frac{2 (\frac{x}{2} \cdot \ln (5))}{\ln (5)}$

Divide $2 (\frac{x}{2})$ by $1$ .

$f^{- 1} \cdot 5^{\frac{x}{2}} = 2 (\frac{x}{2})$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} \cdot 5^{\frac{x}{2}} = 2 (\frac{x}{2})$

Rewrite the expression.

$f^{- 1} \cdot 5^{\frac{x}{2}} = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{2 \ln (x)}{\ln (5)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{2 \ln (x)}{\ln (5)}) = 5^{\frac{\frac{2 \ln (x)}{\ln (5)}}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{2 \ln (x)}{\ln (5)}) = 5^{\frac{2 \ln (x)}{\ln (5)} \cdot \frac{1}{2}}$

Cancel the common factor of $2$ .

Factor $2$ out of $2 \ln (x)$ .

$f (\frac{2 \ln (x)}{\ln (5)}) = 5^{\frac{2 (\ln (x))}{\ln (5)} \cdot \frac{1}{2}}$

Cancel the common factor.

$f (\frac{2 \ln (x)}{\ln (5)}) = 5^{\frac{2 \ln (x)}{\ln (5)} \cdot \frac{1}{2}}$

Rewrite the expression.

$f (\frac{2 \ln (x)}{\ln (5)}) = 5^{\frac{\ln (x)}{\ln (5)}}$

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

$f (\frac{2 \ln (x)}{\ln (5)}) = 5^{\log_{5} (x)}$

Exponentiation and log are inverse functions.

$f (\frac{2 \ln (x)}{\ln (5)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{2 \ln (x)}{\ln (5)}$ is the inverse of $f (x) = 5^{\frac{x}{2}}$ .

$f^{- 1} (x) = \frac{2 \ln (x)}{\ln (5)}$

Find the Inverse y=5^(x/2)

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