Solve for x 2^(x+1)+2^(x-1)=5

$2^{x + 1} + 2^{x - 1} = 5$

Factor out $2^{x - 1}$ from the expression.

$2^{x - 1} (4 + 1)$

Add $4$ and $1$ .

$2^{x - 1} \cdot 5 = 5$

Move $5$ to the left of $2^{x - 1}$ .

$5 \cdot 2^{x - 1} = 5$

Divide each term in $(5 \cdot 2^{x - 1}) = 5$ by $5$ and simplify.

Divide each term in $(5 \cdot 2^{x - 1}) = 5$ by $5$ .

$\frac{(5 \cdot 2^{x - 1})}{5} = \frac{5}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 \cdot 2^{x - 1}}{5} = \frac{5}{5}$

Divide $2^{x - 1}$ by $1$ .

$2^{x - 1} = \frac{5}{5}$

Simplify the right side.

Divide $5$ by $5$ .

$2^{x - 1} = 1$

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

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

Expand $\ln (2^{x - 1})$ by moving $x - 1$ outside the logarithm.

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

Simplify the left side.

Simplify $(x - 1) \ln (2)$ .

Apply the distributive property.

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

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

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

Simplify the right side.

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

$x \ln (2) - \ln (2) = 0$

Add $\ln (2)$ to both sides of the equation.

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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

Cancel the common factor.

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

Rewrite the expression.

$x = 1$

Solve for x 2^(x+1)+2^(x-1)=5

Related Questions

Support LCMGCF.com by disabling your adblocker.