Find the Inverse y=((5^x)/2)^(1/2)

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as ${(\frac{5^{y}}{2})}^{\frac{1}{2}} = x$ .

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

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

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

Expand the left side.

Expand $\ln ({(\frac{5^{y}}{2})}^{\frac{1}{2}})$ by moving $\frac{1}{2}$ outside the logarithm.

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

Rewrite $\ln (\frac{5^{y}}{2})$ as $\ln (5^{y}) - \ln (2)$ .

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

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

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

Simplify the left side.

Simplify $\frac{1}{2} (y \ln (5) - \ln (2))$ .

Apply the distributive property.

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

Multiply $\frac{1}{2} (y \ln (5))$ .

Combine $y$ and $\frac{1}{2}$ .

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

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

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

Combine $\frac{1}{2}$ and $\ln (2)$ .

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

Multiply each term in $\frac{y \ln (5)}{2} - \frac{\ln (2)}{2} = \ln (x)$ by $2$ to eliminate the fractions.

Multiply each term in $\frac{y \ln (5)}{2} - \frac{\ln (2)}{2} = \ln (x)$ by $2$ .

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{\ln (2)}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

$y \ln (5) - \ln (2) = \ln (x) \cdot 2$

Simplify the right side.

Move $2$ to the left of $\ln (x)$ .

$y \ln (5) - \ln (2) = 2 \ln (x)$

Move all the terms containing a logarithm to the left side of the equation.

$y \ln (5) - \ln (2) - 2 \ln (x) = 0$

Move all terms not containing $y$ to the right side of the equation.

Add $\ln (2)$ to both sides of the equation.

$y \ln (5) - 2 \ln (x) = \ln (2)$

Add $2 \ln (x)$ to both sides of the equation.

$y \ln (5) = \ln (2) + 2 \ln (x)$

Divide each term in $y \ln (5) = \ln (2) + 2 \ln (x)$ by $\ln (5)$ and simplify.

Divide each term in $y \ln (5) = \ln (2) + 2 \ln (x)$ by $\ln (5)$ .

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the product rule to $\frac{5^{x}}{2}$ .

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

Multiply the exponents in ${(5^{x})}^{\frac{1}{2}}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $x$ and $\frac{1}{2}$ .

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

Simplify $2 \ln (\frac{5^{\frac{x}{2}}}{2^{\frac{1}{2}}})$ by moving $2$ inside the logarithm.

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

Apply the product rule to $\frac{5^{\frac{x}{2}}}{2^{\frac{1}{2}}}$ .

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

Multiply the exponents in ${(5^{\frac{x}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the denominator.

Multiply the exponents in ${(2^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

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

Simplify the numerator.

Combine the numerators over the common denominator.

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

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

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

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

Exponentiation and log are inverse functions.

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

Reduce the expression by cancelling the common factors.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x^{2}$ by $1$ .

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

Multiply the exponents in ${(x^{2})}^{\frac{1}{2}}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse y=((5^x)/2)^(1/2)

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