Find the Inverse y=(e^(x^8))/2

$y = \frac{e^{x^{8}}}{2}$

Interchange the variables.

$x = \frac{e^{y^{8}}}{2}$

Solve for $y$ .

Rewrite the equation as $\frac{e^{y^{8}}}{2} = x$ .

$\frac{e^{y^{8}}}{2} = x$

Multiply both sides of the equation by $2$ .

$2 \frac{e^{y^{8}}}{2} = 2 x$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{e^{y^{8}}}{2} = 2 x$

Rewrite the expression.

$e^{y^{8}} = 2 x$

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

$\ln (e^{y^{8}}) = \ln (2 x)$

Expand the left side.

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

$y^{8} \ln (e) = \ln (2 x)$

The natural logarithm of $e$ is $1$ .

$y^{8} \cdot 1 = \ln (2 x)$

Multiply $y^{8}$ by $1$ .

$y^{8} = \ln (2 x)$

Take the 8th root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt[8]{\ln (2 x)}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt[8]{\ln (2 x)}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt[8]{\ln (2 x)}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt[8]{\ln (2 x)}$

$y = - \sqrt[8]{\ln (2 x)}$

$y = \sqrt[8]{\ln (2 x)}$

$y = - \sqrt[8]{\ln (2 x)}$

$y = \sqrt[8]{\ln (2 x)}$

$y = - \sqrt[8]{\ln (2 x)}$

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

$f^{- 1} (x) = \sqrt[8]{\ln (2 x)}, - \sqrt[8]{\ln (2 x)}$

Verify if $f^{- 1} (x) = \sqrt[8]{\ln (2 x)}, - \sqrt[8]{\ln (2 x)}$ is the inverse of $f (x) = \frac{e^{x^{8}}}{2}$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \frac{e^{x^{8}}}{2}$ and $f^{- 1} (x) = \sqrt[8]{\ln (2 x)}, - \sqrt[8]{\ln (2 x)}$ and compare them.

Find the range of $f (x) = \frac{e^{x^{8}}}{2}$ .

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[\frac{1}{2}, \infty)$

Find the domain of $\sqrt[8]{\ln (2 x)}$ .

Set the argument in $\ln (2 x)$ greater than $0$ to find where the expression is defined.

$2 x > 0$

Divide each term in $2 x > 0$ by $2$ and simplify.

Divide each term in $2 x > 0$ by $2$ .

$\frac{2 x}{2} > \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} > \frac{0}{2}$

Divide $x$ by $1$ .

$x > \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x > 0$

Set the radicand in $\sqrt[8]{\ln (2 x)}$ greater than or equal to $0$ to find where the expression is defined.

$\ln (2 x) \geq 0$

Solve for $x$ .

Convert the inequality to an equality.

$\ln (2 x) = 0$

Solve the equation.

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (2 x)} = e^{0}$

Rewrite $\ln (2 x) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = 2 x$

Solve for $x$ .

Rewrite the equation as $2 x = e^{0}$ .

$2 x = e^{0}$

Anything raised to $0$ is $1$ .

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Find the domain of $\ln (2 x)$ .

Set the argument in $\ln (2 x)$ greater than $0$ to find where the expression is defined.

$2 x > 0$

Divide each term in $2 x > 0$ by $2$ and simplify.

Divide each term in $2 x > 0$ by $2$ .

$\frac{2 x}{2} > \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} > \frac{0}{2}$

Divide $x$ by $1$ .

$x > \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x > 0$

The domain is all values of $x$ that make the expression defined.

$(0, \infty)$

The solution consists of all of the true intervals.

$x \geq \frac{1}{2}$

The domain is all values of $x$ that make the expression defined.

$[\frac{1}{2}, \infty)$

Find the domain of $f (x) = \frac{e^{x^{8}}}{2}$ .

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Find the range of the inverse.

Find the range of $\sqrt[8]{\ln (2 x)}$ .

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[0, \infty)$

Find the range of $- \sqrt[8]{\ln (2 x)}$ .

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, 0]$

The union consists of all of the elements that are contained in each interval.

$(- \infty, \infty)$

Since the range of $f^{- 1} (x) = \sqrt[8]{\ln (2 x)}, - \sqrt[8]{\ln (2 x)}$ is not equal to the domain of $f (x) = \frac{e^{x^{8}}}{2}$ , then $f^{- 1} (x) = \sqrt[8]{\ln (2 x)}, - \sqrt[8]{\ln (2 x)}$ is not an inverse of $f (x) = \frac{e^{x^{8}}}{2}$ .

There is no inverse

Find the Inverse y=(e^(x^8))/2

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