Find the Inverse f(x) = cube root of 5^x+9

$f (x) = \sqrt[3]{5^{x}} + 9$

Step 1

Write $f (x) = \sqrt[3]{5^{x}} + 9$ as an equation.

$y = \sqrt[3]{5^{x}} + 9$

Step 2

Interchange the variables.

$x = \sqrt[3]{5^{y}} + 9$

Step 3

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{5^{y}} + 9 = x$ .

$\sqrt[3]{5^{y}} + 9 = x$

Subtract $9$ from both sides of the equation.

$\sqrt[3]{5^{y}} = x - 9$

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{5^{y}}}^{3} = {(x - 9)}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{5^{y}}$ as $5^{\frac{y}{3}}$ .

${(5^{\frac{y}{3}})}^{3} = {(x - 9)}^{3}$

Simplify the left side.

Multiply the exponents in ${(5^{\frac{y}{3}})}^{3}$ .

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

$5^{\frac{y}{3} \cdot 3} = {(x - 9)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$5^{\frac{y}{3} \cdot 3} = {(x - 9)}^{3}$

Rewrite the expression.

$5^{y} = {(x - 9)}^{3}$

Simplify the right side.

Simplify ${(x - 9)}^{3}$ .

Use the Binomial Theorem.

$5^{y} = x^{3} + 3 x^{2} \cdot - 9 + 3 x {(- 9)}^{2} + {(- 9)}^{3}$

Simplify each term.

Multiply $- 9$ by $3$ .

$5^{y} = x^{3} - 27 x^{2} + 3 x {(- 9)}^{2} + {(- 9)}^{3}$

Raise $- 9$ to the power of $2$ .

$5^{y} = x^{3} - 27 x^{2} + 3 x \cdot 81 + {(- 9)}^{3}$

Multiply $81$ by $3$ .

$5^{y} = x^{3} - 27 x^{2} + 243 x + {(- 9)}^{3}$

Raise $- 9$ to the power of $3$ .

$5^{y} = x^{3} - 27 x^{2} + 243 x - 729$

Solve for $y$ .

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

$\ln (5^{y}) = \ln (x^{3} - 27 x^{2} + 243 x - 729)$

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

$y \ln (5) = \ln (x^{3} - 27 x^{2} + 243 x - 729)$

Divide each term in $y \ln (5) = \ln (x^{3} - 27 x^{2} + 243 x - 729)$ by $\ln (5)$ and simplify.

Divide each term in $y \ln (5) = \ln (x^{3} - 27 x^{2} + 243 x - 729)$ by $\ln (5)$ .

$\frac{y \ln (5)}{\ln (5)} = \frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{y \ln (5)}{\ln (5)} = \frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}$

Divide $y$ by $1$ .

$y = \frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}$

Step 4

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

$f^{- 1} (x) = \frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}$

Step 5

Verify if $f^{- 1} (x) = \frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}$ is the inverse of $f (x) = \sqrt[3]{5^{x}} + 9$ .

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 $f^{- 1} (\sqrt[3]{5^{x}} + 9)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[3]{5^{x}} + 9) = \frac{\ln ({(\sqrt[3]{5^{x}} + 9)}^{3} - 27 {(\sqrt[3]{5^{x}} + 9)}^{2} + 243 (\sqrt[3]{5^{x}} + 9) - 729)}{\ln (5)}$

Simplify the numerator.

Simplify each term.

Apply the distributive property.

$f^{- 1} (\sqrt[3]{5^{x}} + 9) = \frac{\ln ({(\sqrt[3]{5^{x}} + 9)}^{3} - 27 {(\sqrt[3]{5^{x}} + 9)}^{2} + 243 \sqrt[3]{5^{x}} + 243 \cdot 9 - 729)}{\ln (5)}$

Multiply $243$ by $9$ .

$f^{- 1} (\sqrt[3]{5^{x}} + 9) = \frac{\ln ({(\sqrt[3]{5^{x}} + 9)}^{3} - 27 {(\sqrt[3]{5^{x}} + 9)}^{2} + 243 \sqrt[3]{5^{x}} + 2187 - 729)}{\ln (5)}$

Subtract $729$ from $2187$ .

$f^{- 1} (\sqrt[3]{5^{x}} + 9) = \frac{\ln ({(\sqrt[3]{5^{x}} + 9)}^{3} - 27 {(\sqrt[3]{5^{x}} + 9)}^{2} + 243 \sqrt[3]{5^{x}} + 1458)}{\ln (5)}$

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

Set up the composite result function.

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

Evaluate $f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}) = \sqrt[3]{5^{\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}}} + 9$

Simplify each term.

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

$f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}) = \sqrt[3]{5^{\log_{5} (x^{3} - 27 x^{2} + 243 x - 729)}} + 9$

Exponentiation and log are inverse functions.

$f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}) = \sqrt[3]{x^{3} - 27 x^{2} + 243 x - 729} + 9$

Make each term match the terms from the binomial theorem formula.

$f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}) = \sqrt[3]{x^{3} - 3 \cdot (9 x^{2}) + 3 \cdot (9^{2} x) - 1 \cdot 9^{3}} + 9$

Factor $x^{3} - 3 \cdot 9 x^{2} + 3 \cdot 9^{2} x - 1 \cdot 9^{3}$ using the binomial theorem.

$f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}) = \sqrt[3]{{(x - 9)}^{3}} + 9$

Pull terms out from under the radical, assuming real numbers.

$f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}) = x - 9 + 9$

Combine the opposite terms in $x - 9 + 9$ .

Add $- 9$ and $9$ .

$f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}) = x + 0$

Add $x$ and $0$ .

$f (\frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}$ is the inverse of $f (x) = \sqrt[3]{5^{x}} + 9$ .

$f^{- 1} (x) = \frac{\ln (x^{3} - 27 x^{2} + 243 x - 729)}{\ln (5)}$

Find the Inverse f(x) = cube root of 5^x+9

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