Algebra Examples

$e^{x} + e^{- x} = 3$

Step 1

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

$e^{x} + {(e^{x})}^{- 1} = 3$

Step 2

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

$u + u^{- 1} = 3$

Step 3

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

$u + \frac{1}{u} = 3$

Step 4

Solve for $u$ .

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Step 4.1

Find the LCD of the terms in the equation.

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Step 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 4.1.2

The LCM of one and any expression is the expression.

$u$

Step 4.2

Multiply each term in $u + \frac{1}{u} = 3$ by $u$ to eliminate the fractions.

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Step 4.2.1

Multiply each term in $u + \frac{1}{u} = 3$ by $u$ .

$u \cdot u + \frac{1}{u} u = 3 u$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Multiply $u$ by $u$ .

$u^{2} + \frac{1}{u} u = 3 u$

Step 4.2.2.1.2

Cancel the common factor of $u$ .

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Step 4.2.2.1.2.1

Cancel the common factor.

$u^{2} + \frac{1}{u} u = 3 u$

Step 4.2.2.1.2.2

Rewrite the expression.

$u^{2} + 1 = 3 u$

Step 4.3

Solve the equation.

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Step 4.3.1

Subtract $3 u$ from both sides of the equation.

$u^{2} + 1 - 3 u = 0$

Step 4.3.2

Use the quadratic formula to find the solutions.

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

Step 4.3.3

Substitute the values $a = 1$ , $b = - 3$ , and $c = 1$ into the quadratic formula and solve for $u$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 4.3.4

Simplify.

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Step 4.3.4.1

Simplify the numerator.

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Step 4.3.4.1.1

Raise $- 3$ to the power of $2$ .

$u = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 4.3.4.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 4.3.4.1.2.1

Multiply $- 4$ by $1$ .

$u = \frac{3 \pm \sqrt{9 - 4 \cdot 1}}{2 \cdot 1}$

Step 4.3.4.1.2.2

Multiply $- 4$ by $1$ .

$u = \frac{3 \pm \sqrt{9 - 4}}{2 \cdot 1}$

Step 4.3.4.1.3

Subtract $4$ from $9$ .

$u = \frac{3 \pm \sqrt{5}}{2 \cdot 1}$

Step 4.3.4.2

Multiply $2$ by $1$ .

$u = \frac{3 \pm \sqrt{5}}{2}$

Step 4.3.5

The final answer is the combination of both solutions.

$u = \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}$

Step 5

Substitute $\frac{3 + \sqrt{5}}{2}$ for $u$ in $u = e^{x}$ .

$\frac{3 + \sqrt{5}}{2} = e^{x}$

Step 6

Solve $\frac{3 + \sqrt{5}}{2} = e^{x}$ .

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Step 6.1

Rewrite the equation as $e^{x} = \frac{3 + \sqrt{5}}{2}$ .

$e^{x} = \frac{3 + \sqrt{5}}{2}$

Step 6.2

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

$\ln (e^{x}) = \ln (\frac{3 + \sqrt{5}}{2})$

Step 6.3

Expand the left side.

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Step 6.3.1

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

$x \ln (e) = \ln (\frac{3 + \sqrt{5}}{2})$

Step 6.3.2

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

$x \cdot 1 = \ln (\frac{3 + \sqrt{5}}{2})$

Step 6.3.3

Multiply $x$ by $1$ .

$x = \ln (\frac{3 + \sqrt{5}}{2})$

Step 7

Substitute $\frac{3 - \sqrt{5}}{2}$ for $u$ in $u = e^{x}$ .

$\frac{3 - \sqrt{5}}{2} = e^{x}$

Step 8

Solve $\frac{3 - \sqrt{5}}{2} = e^{x}$ .

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Step 8.1

Rewrite the equation as $e^{x} = \frac{3 - \sqrt{5}}{2}$ .

$e^{x} = \frac{3 - \sqrt{5}}{2}$

Step 8.2

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

$\ln (e^{x}) = \ln (\frac{3 - \sqrt{5}}{2})$

Step 8.3

Expand the left side.

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Step 8.3.1

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

$x \ln (e) = \ln (\frac{3 - \sqrt{5}}{2})$

Step 8.3.2

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

$x \cdot 1 = \ln (\frac{3 - \sqrt{5}}{2})$

Step 8.3.3

Multiply $x$ by $1$ .

$x = \ln (\frac{3 - \sqrt{5}}{2})$

Step 9

List the solutions that makes the equation true.

$x = \ln (\frac{3 + \sqrt{5}}{2}), \ln (\frac{3 - \sqrt{5}}{2})$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = \ln (\frac{3 + \sqrt{5}}{2}), \ln (\frac{3 - \sqrt{5}}{2})$

Decimal Form:

$x = 0.96242365 \dots, - 0.96242365 \dots$