Solve for x e^x+e^(-x)=4

$e^{x} + e^{- x} = 4$

Step 1

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

$e^{x} + {(e^{x})}^{- 1} = 4$

Step 2

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

$u + u^{- 1} = 4$

Step 3

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

$u + \frac{1}{u} = 4$

Step 4

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.

$1, u, 1$

The LCM of one and any expression is the expression.

$u$

Multiply each term in $u + \frac{1}{u} = 4$ by $u$ to eliminate the fractions.

Multiply each term in $u + \frac{1}{u} = 4$ by $u$ .

$u \cdot u + \frac{1}{u} u = 4 u$

Simplify the left side.

Simplify each term.

Multiply $u$ by $u$ .

$u^{2} + \frac{1}{u} u = 4 u$

Cancel the common factor of $u$ .

Cancel the common factor.

$u^{2} + \frac{1}{u} u = 4 u$

Rewrite the expression.

$u^{2} + 1 = 4 u$

Solve the equation.

Subtract $4 u$ from both sides of the equation.

$u^{2} + 1 - 4 u = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $u$ .

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

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $16$ .

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

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$u = \frac{4 \pm \sqrt{4 (3)}}{2 \cdot 1}$

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

$u = \frac{4 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$u = \frac{4 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$u = \frac{4 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{3}}{2}$ .

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

The final answer is the combination of both solutions.

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

Step 5

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

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

Step 6

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

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

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

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

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

Expand the left side.

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

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

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

$x \cdot 1 = \ln (2 + \sqrt{3})$

Multiply $x$ by $1$ .

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

Step 7

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

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

Step 8

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

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

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

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

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

Expand the left side.

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

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

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

$x \cdot 1 = \ln (2 - \sqrt{3})$

Multiply $x$ by $1$ .

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

Step 9

List the solutions that makes the equation true.

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

Step 10

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 1.31695789 \dots, - 1.31695789 \dots$

Solve for x e^x+e^(-x)=4

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