Precalculus Examples

$\frac{e^{x} + 9 e^{- x}}{2} = 3$

Step 1

Multiply both sides by $2$ .

$\frac{e^{x} + 9 e^{- x}}{2} \cdot 2 = 3 \cdot 2$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Cancel the common factor of $2$ .

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Step 2.1.1.1

Cancel the common factor.

$\frac{e^{x} + 9 e^{- x}}{2} \cdot 2 = 3 \cdot 2$

Step 2.1.1.2

Rewrite the expression.

$e^{x} + 9 e^{- x} = 3 \cdot 2$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $3$ by $2$ .

$e^{x} + 9 e^{- x} = 6$

Step 3

Solve for $x$ .

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Step 3.1

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

$e^{x} + 9 {(e^{x})}^{- 1} = 6$

Step 3.2

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

$u + 9 u^{- 1} = 6$

Step 3.3

Simplify each term.

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Step 3.3.1

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

$u + 9 (\frac{1}{u}) = 6$

Step 3.3.2

Combine $9$ and $\frac{1}{u}$ .

$u + \frac{9}{u} = 6$

Step 3.4

Solve for $u$ .

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Step 3.4.1

Find the LCD of the terms in the equation.

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Step 3.4.1.1

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 3.4.1.2

The LCM of one and any expression is the expression.

$u$

Step 3.4.2

Multiply each term in $u + \frac{9}{u} = 6$ by $u$ to eliminate the fractions.

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Step 3.4.2.1

Multiply each term in $u + \frac{9}{u} = 6$ by $u$ .

$u \cdot u + \frac{9}{u} u = 6 u$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Simplify each term.

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Step 3.4.2.2.1.1

Multiply $u$ by $u$ .

$u^{2} + \frac{9}{u} u = 6 u$

Step 3.4.2.2.1.2

Cancel the common factor of $u$ .

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Step 3.4.2.2.1.2.1

Cancel the common factor.

$u^{2} + \frac{9}{u} u = 6 u$

Step 3.4.2.2.1.2.2

Rewrite the expression.

$u^{2} + 9 = 6 u$

Step 3.4.3

Solve the equation.

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Step 3.4.3.1

Subtract $6 u$ from both sides of the equation.

$u^{2} + 9 - 6 u = 0$

Step 3.4.3.2

Factor using the perfect square rule.

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Step 3.4.3.2.1

Rearrange terms.

$u^{2} - 6 u + 9 = 0$

Step 3.4.3.2.2

Rewrite $9$ as $3^{2}$ .

$u^{2} - 6 u + 3^{2} = 0$

Step 3.4.3.2.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Step 3.4.3.2.4

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 3 + 3^{2} = 0$

Step 3.4.3.2.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 3$ .

${(u - 3)}^{2} = 0$

Step 3.4.3.3

Set the $u - 3$ equal to $0$ .

$u - 3 = 0$

Step 3.4.3.4

Add $3$ to both sides of the equation.

$u = 3$

Step 3.5

Substitute $3$ for $u$ in $u = e^{x}$ .

$3 = e^{x}$

Step 3.6

Solve $3 = e^{x}$ .

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Step 3.6.1

Rewrite the equation as $e^{x} = 3$ .

$e^{x} = 3$

Step 3.6.2

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

$\ln (e^{x}) = \ln (3)$

Step 3.6.3

Expand the left side.

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Step 3.6.3.1

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

$x \ln (e) = \ln (3)$

Step 3.6.3.2

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

$x \cdot 1 = \ln (3)$

Step 3.6.3.3

Multiply $x$ by $1$ .

$x = \ln (3)$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \ln (3)$

Decimal Form:

$x = 1.09861228 \dots$