Solve for x 2^x+2^(-x)=2

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

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

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

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

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

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

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

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 $u + \frac{1}{u} = 2$ by $u$ to eliminate the fractions.

Multiply each term in $u + \frac{1}{u} = 2$ by $u$ .

$u \cdot u + \frac{1}{u} u = 2 u$

Simplify the left side.

Simplify each term.

Multiply $u$ by $u$ .

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

Cancel the common factor of $u$ .

Cancel the common factor.

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

Rewrite the expression.

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

Solve the equation.

Subtract $2 u$ from both sides of the equation.

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

Factor using the perfect square rule.

Rearrange terms.

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

Rewrite $1$ as $1^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 1$ .

${(u - 1)}^{2} = 0$

Set the $u - 1$ equal to $0$ .

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

Substitute $1$ for $u$ in $u = 2^{x}$ .

$1 = 2^{x}$

Solve $1 = 2^{x}$ .

Rewrite the equation as $2^{x} = 1$ .

$2^{x} = 1$

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

$\ln (2^{x}) = \ln (1)$

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

$x \ln (2) = \ln (1)$

Simplify the right side.

The natural logarithm of $1$ is $0$ .

$x \ln (2) = 0$

Divide each term in $x \ln (2) = 0$ by $\ln (2)$ and simplify.

Divide each term in $x \ln (2) = 0$ by $\ln (2)$ .

$\frac{x \ln (2)}{\ln (2)} = \frac{0}{\ln (2)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{0}{\ln (2)}$

Divide $x$ by $1$ .

$x = \frac{0}{\ln (2)}$

Simplify the right side.

Divide $0$ by $\ln (2)$ .

$x = 0$

Solve for x 2^x+2^(-x)=2

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