Find the Inverse fourth root of 7x

$\sqrt[4]{7 x}$

Step 1

Interchange the variables.

$x = \sqrt[4]{7 y}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt[4]{7 y} = x$ .

$\sqrt[4]{7 y} = x$

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $4$ .

${\sqrt[4]{7 y}}^{4} = x^{4}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{7 y}$ as ${(7 y)}^{\frac{1}{4}}$ .

${({(7 y)}^{\frac{1}{4}})}^{4} = x^{4}$

Simplify the left side.

Simplify ${({(7 y)}^{\frac{1}{4}})}^{4}$ .

Multiply the exponents in ${({(7 y)}^{\frac{1}{4}})}^{4}$ .

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

${(7 y)}^{\frac{1}{4} \cdot 4} = x^{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

${(7 y)}^{\frac{1}{4} \cdot 4} = x^{4}$

Rewrite the expression.

${(7 y)}^{1} = x^{4}$

Simplify.

$7 y = x^{4}$

Divide each term in $7 y = x^{4}$ by $7$ and simplify.

Divide each term in $7 y = x^{4}$ by $7$ .

$\frac{7 y}{7} = \frac{x^{4}}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y}{7} = \frac{x^{4}}{7}$

Divide $y$ by $1$ .

$y = \frac{x^{4}}{7}$

Step 3

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

$f^{- 1} (x) = \frac{x^{4}}{7}$

Step 4

Verify if $f^{- 1} (x) = \frac{x^{4}}{7}$ is the inverse of $f (x) = \sqrt[4]{7 x}$ .

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 $f^{- 1} (\sqrt[4]{7 x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[4]{7 x}) = \frac{{(\sqrt[4]{7 x})}^{4}}{7}$

Rewrite ${\sqrt[4]{7 x}}^{4}$ as $7 x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{7 x}$ as ${(7 x)}^{\frac{1}{4}}$ .

$f^{- 1} (\sqrt[4]{7 x}) = \frac{{({(7 x)}^{\frac{1}{4}})}^{4}}{7}$

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

$f^{- 1} (\sqrt[4]{7 x}) = \frac{{(7 x)}^{\frac{1}{4} \cdot 4}}{7}$

Combine $\frac{1}{4}$ and $4$ .

$f^{- 1} (\sqrt[4]{7 x}) = \frac{{(7 x)}^{\frac{4}{4}}}{7}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (\sqrt[4]{7 x}) = \frac{{(7 x)}^{\frac{4}{4}}}{7}$

Rewrite the expression.

$f^{- 1} (\sqrt[4]{7 x}) = \frac{7 x}{7}$

Simplify.

$f^{- 1} (\sqrt[4]{7 x}) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (\sqrt[4]{7 x}) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f^{- 1} (\sqrt[4]{7 x}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{4}}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{4}}{7}) = \sqrt[4]{7 (\frac{x^{4}}{7})}$

Combine $7$ and $\frac{x^{4}}{7}$ .

$f (\frac{x^{4}}{7}) = \sqrt[4]{\frac{7 x^{4}}{7}}$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{7 x^{4}}{7}$ by cancelling the common factors.

Cancel the common factor.

$f (\frac{x^{4}}{7}) = \sqrt[4]{\frac{7 x^{4}}{7}}$

Rewrite the expression.

$f (\frac{x^{4}}{7}) = \sqrt[4]{\frac{x^{4}}{1}}$

Divide $x^{4}$ by $1$ .

$f (\frac{x^{4}}{7}) = \sqrt[4]{x^{4}}$

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

$f (\frac{x^{4}}{7}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{4}}{7}$ is the inverse of $f (x) = \sqrt[4]{7 x}$ .

$f^{- 1} (x) = \frac{x^{4}}{7}$

Find the Inverse fourth root of 7x

Related Questions

Support LCMGCF.com by disabling your adblocker.