Precalculus Examples

\frac{2^{x} - 2^{- x}}{3} = 4

$\frac{2^{x} - 2^{- x}}{3} = 4$

Step 1

Multiply both sides by

3

$3$ .

\frac{2^{x} - 2^{- x}}{3} \cdot 3 = 4 \cdot 3

$\frac{2^{x} - 2^{- x}}{3} \cdot 3 = 4 \cdot 3$

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

3

$3$ .

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Step 2.1.1.1

Cancel the common factor.

$\frac{2^{x} - 2^{- x}}{3} \cdot 3 = 4 \cdot 3$

Step 2.1.1.2

Rewrite the expression.

$2^{x} - 2^{- x} = 4 \cdot 3$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply

$4$ by

$3$ .

$2^{x} - 2^{- x} = 12$

Step 3

Solve for

$x$ .

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Step 3.1

Rewrite

$2^{- x}$ as exponentiation.

$2^{x} - {(2^{x})}^{- 1} = 12$

Step 3.2

Substitute

$u$ for

$2^{x}$ .

$u - u^{- 1} = 12$

Step 3.3

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$u - \frac{1}{u} = 12$

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

$u - \frac{1}{u} = 12$ by

$u$ to eliminate the fractions.

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Step 3.4.2.1

Multiply each term in

$u - \frac{1}{u} = 12$ by

$u$ .

$u \cdot u - \frac{1}{u} u = 12 u$

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

$u$ .

$u^{2} - \frac{1}{u} u = 12 u$

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

Move the leading negative in

$- \frac{1}{u}$ into the numerator.

$u^{2} + \frac{- 1}{u} u = 12 u$

Step 3.4.2.2.1.2.2

Cancel the common factor.

$u^{2} + \frac{- 1}{u} u = 12 u$

Step 3.4.2.2.1.2.3

Rewrite the expression.

$u^{2} - 1 = 12 u$

Step 3.4.3

Solve the equation.

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Step 3.4.3.1

Subtract

$12 u$ from both sides of the equation.

$u^{2} - 1 - 12 u = 0$

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

$a = 1$ ,

$b = - 12$ , and

$c = - 1$ into the quadratic formula and solve for

$u$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

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

$- 12$ to the power of

$2$ .

$u = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

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

$1$ .

$u = \frac{12 \pm \sqrt{144 - 4 \cdot - 1}}{2 \cdot 1}$

Step 3.4.3.4.1.2.2

Multiply

$- 4$ by

$- 1$ .

$u = \frac{12 \pm \sqrt{144 + 4}}{2 \cdot 1}$

Step 3.4.3.4.1.3

Add

$144$ and

$4$ .

$u = \frac{12 \pm \sqrt{148}}{2 \cdot 1}$

Step 3.4.3.4.1.4

Rewrite

$148$ as

$2^{2} \cdot 37$ .

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Step 3.4.3.4.1.4.1

Factor

$4$ out of

$148$ .

$u = \frac{12 \pm \sqrt{4 (37)}}{2 \cdot 1}$

Step 3.4.3.4.1.4.2

Rewrite

$4$ as

$2^{2}$ .

$u = \frac{12 \pm \sqrt{2^{2} \cdot 37}}{2 \cdot 1}$

Step 3.4.3.4.1.5

Pull terms out from under the radical.

$u = \frac{12 \pm 2 \sqrt{37}}{2 \cdot 1}$

Step 3.4.3.4.2

Multiply

$2$ by

$1$ .

$u = \frac{12 \pm 2 \sqrt{37}}{2}$

Step 3.4.3.4.3

Simplify

$\frac{12 \pm 2 \sqrt{37}}{2}$ .

$u = 6 \pm \sqrt{37}$

Step 3.4.3.5

The final answer is the combination of both solutions.

$u = 6 + \sqrt{37}, 6 - \sqrt{37}$

Step 3.5

Substitute

$6 + \sqrt{37}$ for

$u$ in

$u = 2^{x}$ .

$6 + \sqrt{37} = 2^{x}$

Step 3.6

Solve

$6 + \sqrt{37} = 2^{x}$ .

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Step 3.6.1

Rewrite the equation as

$2^{x} = 6 + \sqrt{37}$ .

$2^{x} = 6 + \sqrt{37}$

Step 3.6.2

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

$\ln (2^{x}) = \ln (6 + \sqrt{37})$

Step 3.6.3

Expand

$\ln (2^{x})$ by moving

$x$ outside the logarithm.

$x \ln (2) = \ln (6 + \sqrt{37})$

Step 3.6.4

Divide each term in

$x \ln (2) = \ln (6 + \sqrt{37})$ by

$\ln (2)$ and simplify.

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Step 3.6.4.1

Divide each term in

$x \ln (2) = \ln (6 + \sqrt{37})$ by

$\ln (2)$ .

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (6 + \sqrt{37})}{\ln (2)}$

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.

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (6 + \sqrt{37})}{\ln (2)}$

Step 3.6.4.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{\ln (6 + \sqrt{37})}{\ln (2)}$

Step 3.7

Substitute

$6 - \sqrt{37}$ for

$u$ in

$u = 2^{x}$ .

$6 - \sqrt{37} = 2^{x}$

Step 3.8

Solve

$6 - \sqrt{37} = 2^{x}$ .

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Step 3.8.1

Rewrite the equation as

$2^{x} = 6 - \sqrt{37}$ .

$2^{x} = 6 - \sqrt{37}$

Step 3.8.2

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

$\ln (2^{x}) = \ln (6 - \sqrt{37})$

Step 3.8.3

The equation cannot be solved because

$\ln (6 - \sqrt{37})$ is undefined.

Undefined

Step 3.8.4

There is no solution for

$2^{x} = 6 - \sqrt{37}$

No solution

Step 3.9

List the solutions that makes the equation true.

$x = \frac{\ln (6 + \sqrt{37})}{\ln (2)}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (6 + \sqrt{37})}{\ln (2)}$

Decimal Form:

$x = 3.59487843 \dots$