Solve for x -3^(-x)-6=-3^x+10

$- 3^{- x} - 6 = - 3^{x} + 10$

Step 1

Add $3^{x}$ to both sides of the equation.

$- 3^{- x} - 6 + 3^{x} = 10$

Step 2

Rewrite $3^{- x}$ as exponentiation.

$- {(3^{x})}^{- 1} - 6 + 3^{x} = 10$

Step 3

Substitute $u$ for $3^{x}$ .

$- u^{- 1} - 6 + u = 10$

Step 4

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

$- \frac{1}{u} - 6 + u = 10$

Step 5

Move $- 6$ .

$- \frac{1}{u} + u - 6 = 10$

Step 6

Reorder $- \frac{1}{u}$ and $u$ .

$u - \frac{1}{u} - 6 = 10$

Step 7

Solve for $u$ .

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, u, 1, 1$

The LCM of one and any expression is the expression.

$u$

Multiply each term in $u - \frac{1}{u} - 6 = 10$ by $u$ to eliminate the fractions.

Multiply each term in $u - \frac{1}{u} - 6 = 10$ by $u$ .

$u \cdot u - \frac{1}{u} u - 6 u = 10 u$

Simplify the left side.

Simplify each term.

Multiply $u$ by $u$ .

$u^{2} - \frac{1}{u} u - 6 u = 10 u$

Cancel the common factor of $u$ .

Move the leading negative in $- \frac{1}{u}$ into the numerator.

$u^{2} + \frac{- 1}{u} u - 6 u = 10 u$

Cancel the common factor.

$u^{2} + \frac{- 1}{u} u - 6 u = 10 u$

Rewrite the expression.

$u^{2} - 1 - 6 u = 10 u$

Solve the equation.

Move all terms containing $u$ to the left side of the equation.

Subtract $10 u$ from both sides of the equation.

$u^{2} - 1 - 6 u - 10 u = 0$

Subtract $10 u$ from $- 6 u$ .

$u^{2} - 1 - 16 u = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 16$ , and $c = - 1$ into the quadratic formula and solve for $u$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 16$ to the power of $2$ .

$u = \frac{16 \pm \sqrt{256 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$u = \frac{16 \pm \sqrt{256 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$u = \frac{16 \pm \sqrt{256 + 4}}{2 \cdot 1}$

Add $256$ and $4$ .

$u = \frac{16 \pm \sqrt{260}}{2 \cdot 1}$

Rewrite $260$ as $2^{2} \cdot 65$ .

Factor $4$ out of $260$ .

$u = \frac{16 \pm \sqrt{4 (65)}}{2 \cdot 1}$

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

$u = \frac{16 \pm \sqrt{2^{2} \cdot 65}}{2 \cdot 1}$

Pull terms out from under the radical.

$u = \frac{16 \pm 2 \sqrt{65}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$u = \frac{16 \pm 2 \sqrt{65}}{2}$

Simplify $\frac{16 \pm 2 \sqrt{65}}{2}$ .

$u = 8 \pm \sqrt{65}$

The final answer is the combination of both solutions.

$u = 8 + \sqrt{65}, 8 - \sqrt{65}$

Step 8

Substitute $8 + \sqrt{65}$ for $u$ in $u = 3^{x}$ .

$8 + \sqrt{65} = 3^{x}$

Step 9

Solve $8 + \sqrt{65} = 3^{x}$ .

Rewrite the equation as $3^{x} = 8 + \sqrt{65}$ .

$3^{x} = 8 + \sqrt{65}$

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

$\ln (3^{x}) = \ln (8 + \sqrt{65})$

Expand $\ln (3^{x})$ by moving $x$ outside the logarithm.

$x \ln (3) = \ln (8 + \sqrt{65})$

Divide each term in $x \ln (3) = \ln (8 + \sqrt{65})$ by $\ln (3)$ and simplify.

Divide each term in $x \ln (3) = \ln (8 + \sqrt{65})$ by $\ln (3)$ .

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (8 + \sqrt{65})}{\ln (3)}$

Simplify the left side.

Cancel the common factor of $\ln (3)$ .

Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (8 + \sqrt{65})}{\ln (3)}$

Divide $x$ by $1$ .

$x = \frac{\ln (8 + \sqrt{65})}{\ln (3)}$

Step 10

Substitute $8 - \sqrt{65}$ for $u$ in $u = 3^{x}$ .

$8 - \sqrt{65} = 3^{x}$

Step 11

Solve $8 - \sqrt{65} = 3^{x}$ .

Rewrite the equation as $3^{x} = 8 - \sqrt{65}$ .

$3^{x} = 8 - \sqrt{65}$

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

$\ln (3^{x}) = \ln (8 - \sqrt{65})$

The equation cannot be solved because $\ln (8 - \sqrt{65})$ is undefined.

Undefined

There is no solution for $3^{x} = 8 - \sqrt{65}$

No solution

Step 12

List the solutions that makes the equation true.

$x = \frac{\ln (8 + \sqrt{65})}{\ln (3)}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (8 + \sqrt{65})}{\ln (3)}$

Decimal Form:

$x = 2.52725398 \dots$

Solve for x -3^(-x)-6=-3^x+10

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