Find the Inverse cube root of 8^x

$\sqrt[3]{8^{x}}$

Step 1

Interchange the variables.

$x = \sqrt[3]{8^{y}}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{8^{y}} = x$ .

$\sqrt[3]{8^{y}} = x$

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

${\sqrt[3]{8^{y}}}^{3} = x^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{8^{y}}$ as $8^{\frac{y}{3}}$ .

${(8^{\frac{y}{3}})}^{3} = x^{3}$

Simplify the left side.

Multiply the exponents in ${(8^{\frac{y}{3}})}^{3}$ .

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

$8^{\frac{y}{3} \cdot 3} = x^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8^{\frac{y}{3} \cdot 3} = x^{3}$

Rewrite the expression.

$8^{y} = x^{3}$

Solve for $y$ .

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

$\ln (8^{y}) = \ln (x^{3})$

Expand $\ln (8^{y})$ by moving $y$ outside the logarithm.

$y \ln (8) = \ln (x^{3})$

Divide each term in $y \ln (8) = \ln (x^{3})$ by $\ln (8)$ and simplify.

Divide each term in $y \ln (8) = \ln (x^{3})$ by $\ln (8)$ .

$\frac{y \ln (8)}{\ln (8)} = \frac{\ln (x^{3})}{\ln (8)}$

Simplify the left side.

Cancel the common factor of $\ln (8)$ .

Cancel the common factor.

$\frac{y \ln (8)}{\ln (8)} = \frac{\ln (x^{3})}{\ln (8)}$

Divide $y$ by $1$ .

$y = \frac{\ln (x^{3})}{\ln (8)}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (x^{3})}{\ln (8)}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (x^{3})}{\ln (8)}$ is the inverse of $f (x) = \sqrt[3]{8^{x}}$ .

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 $f^{- 1} (\sqrt[3]{8^{x}})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[3]{8^{x}}) = \frac{\ln ({(\sqrt[3]{8^{x}})}^{3})}{\ln (8)}$

Rewrite ${\sqrt[3]{8^{x}}}^{3}$ as $8^{x}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{8^{x}}$ as $8^{\frac{x}{3}}$ .

$f^{- 1} (\sqrt[3]{8^{x}}) = \frac{\ln ({(8^{\frac{x}{3}})}^{3})}{\ln (8)}$

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

$f^{- 1} (\sqrt[3]{8^{x}}) = \frac{\ln (8^{\frac{x}{3} \cdot 3})}{\ln (8)}$

Combine $\frac{x}{3}$ and $3$ .

$f^{- 1} (\sqrt[3]{8^{x}}) = \frac{\ln (8^{\frac{x \cdot 3}{3}})}{\ln (8)}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{8^{x}}) = \frac{\ln (8^{\frac{x \cdot 3}{3}})}{\ln (8)}$

Divide $x$ by $1$ .

$f^{- 1} (\sqrt[3]{8^{x}}) = \frac{\ln (8^{x})}{\ln (8)}$

Expand $\ln (8^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} (\sqrt[3]{8^{x}}) = \frac{x \ln (8)}{\ln (8)}$

Cancel the common factor of $\ln (8)$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{8^{x}}) = \frac{x \ln (8)}{\ln (8)}$

Divide $x$ by $1$ .

$f^{- 1} (\sqrt[3]{8^{x}}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\ln (x^{3})}{\ln (8)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x^{3})}{\ln (8)}) = \sqrt[3]{8^{\frac{\ln (x^{3})}{\ln (8)}}}$

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

$f (\frac{\ln (x^{3})}{\ln (8)}) = \sqrt[3]{8^{\log_{8} (x^{3})}}$

Exponentiation and log are inverse functions.

$f (\frac{\ln (x^{3})}{\ln (8)}) = \sqrt[3]{x^{3}}$

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

$f (\frac{\ln (x^{3})}{\ln (8)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x^{3})}{\ln (8)}$ is the inverse of $f (x) = \sqrt[3]{8^{x}}$ .

$f^{- 1} (x) = \frac{\ln (x^{3})}{\ln (8)}$

Find the Inverse cube root of 8^x

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