Precalculus Examples

$y = \frac{e^{x} + e^{- x}}{2}$

Step 1

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

$\frac{e^{x} + e^{- x}}{2} = y$

Step 2

Multiply both sides by $2$ .

$\frac{e^{x} + e^{- x}}{2} \cdot 2 = y \cdot 2$

Step 3

Simplify.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Cancel the common factor of $2$ .

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Step 3.1.1.1

Cancel the common factor.

$\frac{e^{x} + e^{- x}}{2} \cdot 2 = y \cdot 2$

Step 3.1.1.2

Rewrite the expression.

$e^{x} + e^{- x} = y \cdot 2$

Step 3.2

Simplify the right side.

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Step 3.2.1

Move $2$ to the left of $y$ .

$e^{x} + e^{- x} = 2 y$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite $e^{- x}$ as exponentiation.

$e^{x} + {(e^{x})}^{- 1} = 2 y$

Step 4.2

Substitute $u$ for $e^{x}$ .

$u + u^{- 1} = 2 y$

Step 4.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$u + \frac{1}{u} = 2 y$

Step 4.4

Solve for $u$ .

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Step 4.4.1

Find the LCD of the terms in the equation.

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Step 4.4.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, u, 1$

Step 4.4.1.2

The LCM of one and any expression is the expression.

$u$

Step 4.4.2

Multiply each term in $u + \frac{1}{u} = 2 y$ by $u$ to eliminate the fractions.

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Step 4.4.2.1

Multiply each term in $u + \frac{1}{u} = 2 y$ by $u$ .

$u \cdot u + \frac{1}{u} u = 2 y u$

Step 4.4.2.2

Simplify the left side.

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Step 4.4.2.2.1

Simplify each term.

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Step 4.4.2.2.1.1

Multiply $u$ by $u$ .

$u^{2} + \frac{1}{u} u = 2 y u$

Step 4.4.2.2.1.2

Cancel the common factor of $u$ .

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Step 4.4.2.2.1.2.1

Cancel the common factor.

$u^{2} + \frac{1}{u} u = 2 y u$

Step 4.4.2.2.1.2.2

Rewrite the expression.

$u^{2} + 1 = 2 y u$

Step 4.4.3

Solve the equation.

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Step 4.4.3.1

Subtract $2 y u$ from both sides of the equation.

$u^{2} + 1 - 2 y u = 0$

Step 4.4.3.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.4.3.3

Substitute the values $a = 1$ , $b = - 2 y$ , and $c = 1$ into the quadratic formula and solve for $u$ .

$\frac{2 y \pm \sqrt{{(- 2 y)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 4.4.3.4

Simplify.

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Step 4.4.3.4.1

Simplify the numerator.

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Step 4.4.3.4.1.1

Rewrite $4 \cdot 1 \cdot 1$ as ${(2 \cdot 1 \cdot 1)}^{2}$ .

$u = \frac{2 y \pm \sqrt{{(- 2 y)}^{2} - {(2 \cdot 1 \cdot 1)}^{2}}}{2 \cdot 1}$

Step 4.4.3.4.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = - 2 y$ and $b = 2 \cdot 1 \cdot 1$ .

$u = \frac{2 y \pm \sqrt{(- 2 y + 2 \cdot 1 \cdot 1) (- 2 y - (2 \cdot 1 \cdot 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.3

Simplify.

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Step 4.4.3.4.1.3.1

Factor $2$ out of $- 2 y + 2 \cdot 1 \cdot 1$ .

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Step 4.4.3.4.1.3.1.1

Factor $2$ out of $- 2 y$ .

$u = \frac{2 y \pm \sqrt{(2 (- y) + 2 \cdot 1 \cdot 1) (- 2 y - (2 \cdot 1 \cdot 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.3.1.2

Factor $2$ out of $2 \cdot 1 \cdot 1$ .

$u = \frac{2 y \pm \sqrt{(2 (- y) + 2 (1 \cdot 1)) (- 2 y - (2 \cdot 1 \cdot 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.3.1.3

Factor $2$ out of $2 (- y) + 2 (1 \cdot 1)$ .

$u = \frac{2 y \pm \sqrt{2 (- y + 1 \cdot 1) (- 2 y - (2 \cdot 1 \cdot 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.3.2

Multiply $1$ by $1$ .

$u = \frac{2 y \pm \sqrt{2 (- y + 1) (- 2 y - (2 \cdot 1 \cdot 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.3.3

Combine exponents.

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Step 4.4.3.4.1.3.3.1

Multiply $2$ by $1$ .

$u = \frac{2 y \pm \sqrt{2 (- y + 1) (- 2 y - (2 \cdot 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.3.3.2

Multiply $2$ by $1$ .

$u = \frac{2 y \pm \sqrt{2 (- y + 1) (- 2 y - 1 \cdot 2)}}{2 \cdot 1}$

Step 4.4.3.4.1.3.3.3

Multiply $- 1$ by $2$ .

$u = \frac{2 y \pm \sqrt{2 (- y + 1) (- 2 y - 2)}}{2 \cdot 1}$

Step 4.4.3.4.1.4

Factor $2$ out of $- 2 y - 2$ .

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Step 4.4.3.4.1.4.1

Factor $2$ out of $- 2 y$ .

$u = \frac{2 y \pm \sqrt{2 (- y + 1) (2 (- y) - 2)}}{2 \cdot 1}$

Step 4.4.3.4.1.4.2

Factor $2$ out of $- 2$ .

$u = \frac{2 y \pm \sqrt{2 (- y + 1) (2 (- y) + 2 (- 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.4.3

Factor $2$ out of $2 (- y) + 2 (- 1)$ .

$u = \frac{2 y \pm \sqrt{2 (- y + 1) (2 (- y - 1))}}{2 \cdot 1}$

$u = \frac{2 y \pm \sqrt{2 (- y + 1) \cdot (2 (- y - 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.5

Multiply $2$ by $2$ .

$u = \frac{2 y \pm \sqrt{4 (- y + 1) (- y - 1)}}{2 \cdot 1}$

Step 4.4.3.4.1.6

Rewrite $4 (- y + 1) (- y - 1)$ as $2^{2} ((- y + 1^{2}) (- y - 1))$ .

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Step 4.4.3.4.1.6.1

Rewrite $4$ as $2^{2}$ .

$u = \frac{2 y \pm \sqrt{2^{2} (- y + 1) (- y - 1)}}{2 \cdot 1}$

Step 4.4.3.4.1.6.2

Rewrite $1$ as $1^{2}$ .

$u = \frac{2 y \pm \sqrt{2^{2} (- y + 1^{2}) (- y - 1)}}{2 \cdot 1}$

Step 4.4.3.4.1.6.3

Add parentheses.

$u = \frac{2 y \pm \sqrt{2^{2} ((- y + 1^{2}) (- y - 1))}}{2 \cdot 1}$

Step 4.4.3.4.1.7

Pull terms out from under the radical.

$u = \frac{2 y \pm 2 \sqrt{(- y + 1^{2}) (- y - 1)}}{2 \cdot 1}$

Step 4.4.3.4.1.8

One to any power is one.

$u = \frac{2 y \pm 2 \sqrt{(- y + 1) (- y - 1)}}{2 \cdot 1}$

Step 4.4.3.4.2

Multiply $2$ by $1$ .

$u = \frac{2 y \pm 2 \sqrt{(- y + 1) (- y - 1)}}{2}$

Step 4.4.3.4.3

Simplify $\frac{2 y \pm 2 \sqrt{(- y + 1) (- y - 1)}}{2}$ .

$u = y \pm \sqrt{(- y + 1) (- y - 1)}$

Step 4.4.3.5

The final answer is the combination of both solutions.

$u = y + \sqrt{(- y + 1) (- y - 1)}$

$u = y - \sqrt{(- y + 1) (- y - 1)}$

$u = y + \sqrt{(- y + 1) (- y - 1)}$

$u = y - \sqrt{(- y + 1) (- y - 1)}$

$u = y + \sqrt{(- y + 1) (- y - 1)}$

$u = y - \sqrt{(- y + 1) (- y - 1)}$

Step 4.5

Substitute $y + \sqrt{(- y + 1) (- y - 1)}$ for $u$ in $u = e^{x}$ .

$y + \sqrt{(- y + 1) (- y - 1)} = e^{x}$

Step 4.6

Solve $y + \sqrt{(- y + 1) (- y - 1)} = e^{x}$ .

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Step 4.6.1

Rewrite the equation as $e^{x} = y + \sqrt{(- y + 1) (- y - 1)}$ .

$e^{x} = y + \sqrt{(- y + 1) (- y - 1)}$

Step 4.6.2

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

$\ln (e^{x}) = \ln (y + \sqrt{(- y + 1) (- y - 1)})$

Step 4.6.3

Expand the left side.

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Step 4.6.3.1

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

$x \ln (e) = \ln (y + \sqrt{(- y + 1) (- y - 1)})$

Step 4.6.3.2

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

$x \cdot 1 = \ln (y + \sqrt{(- y + 1) (- y - 1)})$

Step 4.6.3.3

Multiply $x$ by $1$ .

$x = \ln (y + \sqrt{(- y + 1) (- y - 1)})$

Step 4.7

Substitute $y - \sqrt{(- y + 1) (- y - 1)}$ for $u$ in $u = e^{x}$ .

$y - \sqrt{(- y + 1) (- y - 1)} = e^{x}$

Step 4.8

Solve $y - \sqrt{(- y + 1) (- y - 1)} = e^{x}$ .

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Step 4.8.1

Rewrite the equation as $e^{x} = y - \sqrt{(- y + 1) (- y - 1)}$ .

$e^{x} = y - \sqrt{(- y + 1) (- y - 1)}$

Step 4.8.2

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

$\ln (e^{x}) = \ln (y - \sqrt{(- y + 1) (- y - 1)})$

Step 4.8.3

Expand the left side.

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Step 4.8.3.1

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

$x \ln (e) = \ln (y - \sqrt{(- y + 1) (- y - 1)})$

Step 4.8.3.2

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

$x \cdot 1 = \ln (y - \sqrt{(- y + 1) (- y - 1)})$

Step 4.8.3.3

Multiply $x$ by $1$ .

$x = \ln (y - \sqrt{(- y + 1) (- y - 1)})$

Step 4.9

List the solutions that makes the equation true.

$x = \ln (y + \sqrt{(- y + 1) (- y - 1)})$

$x = \ln (y - \sqrt{(- y + 1) (- y - 1)})$

$x = \ln (y + \sqrt{(- y + 1) (- y - 1)})$

$x = \ln (y - \sqrt{(- y + 1) (- y - 1)})$