Finite Math Examples

$7^{x} + 7^{- x} = 50$

Step 1

Rewrite $7^{- x}$ as exponentiation.

$7^{x} + {(7^{x})}^{- 1} = 50$

Step 2

Substitute $u$ for $7^{x}$ .

$u + u^{- 1} = 50$

Step 3

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

$u + \frac{1}{u} = 50$

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 $u + \frac{1}{u} = 50$ by $u$ to eliminate the fractions.

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Step 4.2.1

Multiply each term in $u + \frac{1}{u} = 50$ by $u$ .

$u \cdot u + \frac{1}{u} u = 50 u$

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 $u$ .

$u^{2} + \frac{1}{u} u = 50 u$

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.

$u^{2} + \frac{1}{u} u = 50 u$

Step 4.2.2.1.2.2

Rewrite the expression.

$u^{2} + 1 = 50 u$

Step 4.3

Solve the equation.

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Step 4.3.1

Subtract $50 u$ from both sides of the equation.

$u^{2} + 1 - 50 u = 0$

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 $a = 1$ , $b = - 50$ , and $c = 1$ into the quadratic formula and solve for $u$ .

$\frac{50 \pm \sqrt{{(- 50)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

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 $- 50$ to the power of $2$ .

$u = \frac{50 \pm \sqrt{2500 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

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 $1$ .

$u = \frac{50 \pm \sqrt{2500 - 4 \cdot 1}}{2 \cdot 1}$

Step 4.3.4.1.2.2

Multiply $- 4$ by $1$ .

$u = \frac{50 \pm \sqrt{2500 - 4}}{2 \cdot 1}$

Step 4.3.4.1.3

Subtract $4$ from $2500$ .

$u = \frac{50 \pm \sqrt{2496}}{2 \cdot 1}$

Step 4.3.4.1.4

Rewrite $2496$ as $8^{2} \cdot 39$ .

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Step 4.3.4.1.4.1

Factor $64$ out of $2496$ .

$u = \frac{50 \pm \sqrt{64 (39)}}{2 \cdot 1}$

Step 4.3.4.1.4.2

Rewrite $64$ as $8^{2}$ .

$u = \frac{50 \pm \sqrt{8^{2} \cdot 39}}{2 \cdot 1}$

Step 4.3.4.1.5

Pull terms out from under the radical.

$u = \frac{50 \pm 8 \sqrt{39}}{2 \cdot 1}$

Step 4.3.4.2

Multiply $2$ by $1$ .

$u = \frac{50 \pm 8 \sqrt{39}}{2}$

Step 4.3.4.3

Simplify $\frac{50 \pm 8 \sqrt{39}}{2}$ .

$u = 25 \pm 4 \sqrt{39}$

Step 4.3.5

The final answer is the combination of both solutions.

$u = 25 + 4 \sqrt{39}, 25 - 4 \sqrt{39}$

Step 5

Substitute $25 + 4 \sqrt{39}$ for $u$ in $u = 7^{x}$ .

$25 + 4 \sqrt{39} = 7^{x}$

Step 6

Solve $25 + 4 \sqrt{39} = 7^{x}$ .

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Step 6.1

Rewrite the equation as $7^{x} = 25 + 4 \sqrt{39}$ .

$7^{x} = 25 + 4 \sqrt{39}$

Step 6.2

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

$\ln (7^{x}) = \ln (25 + 4 \sqrt{39})$

Step 6.3

Expand $\ln (7^{x})$ by moving $x$ outside the logarithm.

$x \ln (7) = \ln (25 + 4 \sqrt{39})$

Step 6.4

Divide each term in $x \ln (7) = \ln (25 + 4 \sqrt{39})$ by $\ln (7)$ and simplify.

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Step 6.4.1

Divide each term in $x \ln (7) = \ln (25 + 4 \sqrt{39})$ by $\ln (7)$ .

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (25 + 4 \sqrt{39})}{\ln (7)}$

Step 6.4.2

Simplify the left side.

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Step 6.4.2.1

Cancel the common factor of $\ln (7)$ .

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Step 6.4.2.1.1

Cancel the common factor.

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (25 + 4 \sqrt{39})}{\ln (7)}$

Step 6.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (25 + 4 \sqrt{39})}{\ln (7)}$

Step 7

Substitute $25 - 4 \sqrt{39}$ for $u$ in $u = 7^{x}$ .

$25 - 4 \sqrt{39} = 7^{x}$

Step 8

Solve $25 - 4 \sqrt{39} = 7^{x}$ .

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Step 8.1

Rewrite the equation as $7^{x} = 25 - 4 \sqrt{39}$ .

$7^{x} = 25 - 4 \sqrt{39}$

Step 8.2

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

$\ln (7^{x}) = \ln (25 - 4 \sqrt{39})$

Step 8.3

Expand $\ln (7^{x})$ by moving $x$ outside the logarithm.

$x \ln (7) = \ln (25 - 4 \sqrt{39})$

Step 8.4

Divide each term in $x \ln (7) = \ln (25 - 4 \sqrt{39})$ by $\ln (7)$ and simplify.

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Step 8.4.1

Divide each term in $x \ln (7) = \ln (25 - 4 \sqrt{39})$ by $\ln (7)$ .

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (25 - 4 \sqrt{39})}{\ln (7)}$

Step 8.4.2

Simplify the left side.

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Step 8.4.2.1

Cancel the common factor of $\ln (7)$ .

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Step 8.4.2.1.1

Cancel the common factor.

$\frac{x \ln (7)}{\ln (7)} = \frac{\ln (25 - 4 \sqrt{39})}{\ln (7)}$

Step 8.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (25 - 4 \sqrt{39})}{\ln (7)}$

Step 9

List the solutions that makes the equation true.

$x = \frac{\ln (25 + 4 \sqrt{39})}{\ln (7)}, \frac{\ln (25 - 4 \sqrt{39})}{\ln (7)}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (25 + 4 \sqrt{39})}{\ln (7)}, \frac{\ln (25 - 4 \sqrt{39})}{\ln (7)}$

Decimal Form:

$x = 2.01017645 \dots, - 2.01017645 \dots$