Algebra Examples

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

Step 1

Multiply both sides by $2$ .

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

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.

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

Step 2.1.1.2

Rewrite the expression.

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

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $3$ by $2$ .

$2^{x} + 2^{- x} = 6$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite $2^{- x}$ as exponentiation.

$2^{x} + {(2^{x})}^{- 1} = 6$

Step 3.2

Substitute $u$ for $2^{x}$ .

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

Step 3.3

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

$u + \frac{1}{u} = 6$

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 $u + \frac{1}{u} = 6$ by $u$ to eliminate the fractions.

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Step 3.4.2.1

Multiply each term in $u + \frac{1}{u} = 6$ by $u$ .

$u \cdot u + \frac{1}{u} u = 6 u$

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 $u$ .

$u^{2} + \frac{1}{u} u = 6 u$

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.

$u^{2} + \frac{1}{u} u = 6 u$

Step 3.4.2.2.1.2.2

Rewrite the expression.

$u^{2} + 1 = 6 u$

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.

$u^{2} + 1 - 6 u = 0$

Step 3.4.3.2

Use the quadratic formula to find the solutions.

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

Step 3.4.3.3

Substitute the values $a = 1$ , $b = - 6$ , and $c = 1$ into the quadratic formula and solve for $u$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 3.4.3.4

Simplify.

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Step 3.4.3.4.1

Simplify the numerator.

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Step 3.4.3.4.1.1

Raise $- 6$ to the power of $2$ .

$u = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 3.4.3.4.1.2

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

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Step 3.4.3.4.1.2.1

Multiply $- 4$ by $1$ .

$u = \frac{6 \pm \sqrt{36 - 4 \cdot 1}}{2 \cdot 1}$

Step 3.4.3.4.1.2.2

Multiply $- 4$ by $1$ .

$u = \frac{6 \pm \sqrt{36 - 4}}{2 \cdot 1}$

Step 3.4.3.4.1.3

Subtract $4$ from $36$ .

$u = \frac{6 \pm \sqrt{32}}{2 \cdot 1}$

Step 3.4.3.4.1.4

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 3.4.3.4.1.4.1

Factor $16$ out of $32$ .

$u = \frac{6 \pm \sqrt{16 (2)}}{2 \cdot 1}$

Step 3.4.3.4.1.4.2

Rewrite $16$ as $4^{2}$ .

$u = \frac{6 \pm \sqrt{4^{2} \cdot 2}}{2 \cdot 1}$

Step 3.4.3.4.1.5

Pull terms out from under the radical.

$u = \frac{6 \pm 4 \sqrt{2}}{2 \cdot 1}$

Step 3.4.3.4.2

Multiply $2$ by $1$ .

$u = \frac{6 \pm 4 \sqrt{2}}{2}$

Step 3.4.3.4.3

Simplify $\frac{6 \pm 4 \sqrt{2}}{2}$ .

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

Step 3.4.3.5

The final answer is the combination of both solutions.

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

Step 3.5

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

$3 + 2 \sqrt{2} = 2^{x}$

Step 3.6

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

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Step 3.6.1

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

$2^{x} = 3 + 2 \sqrt{2}$

Step 3.6.2

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

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

Step 3.6.3

Expand $\ln (2^{x})$ by moving $x$ outside the logarithm.

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

Step 3.6.4

Divide each term in $x \ln (2) = \ln (3 + 2 \sqrt{2})$ by $\ln (2)$ and simplify.

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Step 3.6.4.1

Divide each term in $x \ln (2) = \ln (3 + 2 \sqrt{2})$ by $\ln (2)$ .

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

Step 3.6.4.2

Simplify the left side.

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Step 3.6.4.2.1

Cancel the common factor of $\ln (2)$ .

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Step 3.6.4.2.1.1

Cancel the common factor.

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

Step 3.6.4.2.1.2

Divide $x$ by $1$ .

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

Step 3.7

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

$3 - 2 \sqrt{2} = 2^{x}$

Step 3.8

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

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Step 3.8.1

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

$2^{x} = 3 - 2 \sqrt{2}$

Step 3.8.2

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

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

Step 3.8.3

Expand $\ln (2^{x})$ by moving $x$ outside the logarithm.

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

Step 3.8.4

Divide each term in $x \ln (2) = \ln (3 - 2 \sqrt{2})$ by $\ln (2)$ and simplify.

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Step 3.8.4.1

Divide each term in $x \ln (2) = \ln (3 - 2 \sqrt{2})$ by $\ln (2)$ .

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

Step 3.8.4.2

Simplify the left side.

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Step 3.8.4.2.1

Cancel the common factor of $\ln (2)$ .

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Step 3.8.4.2.1.1

Cancel the common factor.

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

Step 3.8.4.2.1.2

Divide $x$ by $1$ .

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

Step 3.9

List the solutions that makes the equation true.

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 2.54310660 \dots, - 2.54310660 \dots$