Algebra Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

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

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

Step 2.2

Multiply both sides of the equation by $2$ .

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

Step 2.3

Simplify the left side.

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Step 2.3.1

Cancel the common factor of $2$ .

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Step 2.3.1.1

Cancel the common factor.

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

Step 2.3.1.2

Rewrite the expression.

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

Step 2.4

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

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

Step 2.5

Expand the left side.

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Step 2.5.1

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

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

Step 2.5.2

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

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

Step 2.5.3

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

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

Step 2.6

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

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

Step 2.7

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

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Step 2.7.1

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

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

Step 2.7.2

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

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

Step 2.7.3

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

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

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

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

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

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

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

Step 3

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

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

Step 4

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

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Step 4.1

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.

Step 4.2

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

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Step 4.2.1

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

Interval Notation:

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

Step 4.3

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

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Step 4.3.1

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

$2 x > 0$

Step 4.3.2

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

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Step 4.3.2.1

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

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

Step 4.3.2.2

Simplify the left side.

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Step 4.3.2.2.1

Cancel the common factor of $2$ .

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Step 4.3.2.2.1.1

Cancel the common factor.

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

Step 4.3.2.2.1.2

Divide $x$ by $1$ .

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

Step 4.3.2.3

Simplify the right side.

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Step 4.3.2.3.1

Divide $0$ by $2$ .

$x > 0$

Step 4.3.3

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

$\ln (2 x) \geq 0$

Step 4.3.4

Solve for $x$ .

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Step 4.3.4.1

Convert the inequality to an equality.

$\ln (2 x) = 0$

Step 4.3.4.2

Solve the equation.

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Step 4.3.4.2.1

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

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

Step 4.3.4.2.2

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 $b^{y} = x$ .

$e^{0} = 2 x$

Step 4.3.4.2.3

Solve for $x$ .

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Step 4.3.4.2.3.1

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

$2 x = e^{0}$

Step 4.3.4.2.3.2

Anything raised to $0$ is $1$ .

$2 x = 1$

Step 4.3.4.2.3.3

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

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Step 4.3.4.2.3.3.1

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

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

Step 4.3.4.2.3.3.2

Simplify the left side.

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Step 4.3.4.2.3.3.2.1

Cancel the common factor of $2$ .

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Step 4.3.4.2.3.3.2.1.1

Cancel the common factor.

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

Step 4.3.4.2.3.3.2.1.2

Divide $x$ by $1$ .

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

Step 4.3.4.3

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

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Step 4.3.4.3.1

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

$2 x > 0$

Step 4.3.4.3.2

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

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Step 4.3.4.3.2.1

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

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

Step 4.3.4.3.2.2

Simplify the left side.

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Step 4.3.4.3.2.2.1

Cancel the common factor of $2$ .

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Step 4.3.4.3.2.2.1.1

Cancel the common factor.

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

Step 4.3.4.3.2.2.1.2

Divide $x$ by $1$ .

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

Step 4.3.4.3.2.3

Simplify the right side.

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Step 4.3.4.3.2.3.1

Divide $0$ by $2$ .

$x > 0$

Step 4.3.4.3.3

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

$(0, \infty)$

Step 4.3.4.4

The solution consists of all of the true intervals.

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

Step 4.3.5

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

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

Step 4.4

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

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Step 4.4.1

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.

$(- \infty, \infty)$

Step 4.5

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

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

Step 5