Solve by Factoring 5^(x+1)+5^(x-1)=26

$5^{x + 1} + 5^{x - 1} = 26$

Subtract $26$ from both sides of the equation.

$5^{x + 1} + 5^{x - 1} - 26 = 0$

Factor out $5^{x - 1}$ from the expression.

$5^{x - 1} (25 + 1)$

Add $26$ to both sides of the equation.

$5^{x + 1} + 5^{x - 1} = 26$

Add $25$ and $1$ .

$5^{x - 1} \cdot 26 = 26$

Move $26$ to the left of $5^{x - 1}$ .

$26 \cdot 5^{x - 1} = 26$

Divide each term in $(26 \cdot 5^{x - 1}) = 26$ by $26$ and simplify.

Divide each term in $(26 \cdot 5^{x - 1}) = 26$ by $26$ .

$\frac{(26 \cdot 5^{x - 1})}{26} = \frac{26}{26}$

Simplify the left side.

Cancel the common factor of $26$ .

Cancel the common factor.

$\frac{26 \cdot 5^{x - 1}}{26} = \frac{26}{26}$

Divide $5^{x - 1}$ by $1$ .

$5^{x - 1} = \frac{26}{26}$

Simplify the right side.

Divide $26$ by $26$ .

$5^{x - 1} = 1$

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

$\ln (5^{x - 1}) = \ln (1)$

Expand $\ln (5^{x - 1})$ by moving $x - 1$ outside the logarithm.

$(x - 1) \ln (5) = \ln (1)$

Simplify the left side.

Simplify $(x - 1) \ln (5)$ .

Apply the distributive property.

$x \ln (5) - 1 \ln (5) = \ln (1)$

Rewrite $- 1 \ln (5)$ as $- \ln (5)$ .

$x \ln (5) - \ln (5) = \ln (1)$

Simplify the right side.

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

$x \ln (5) - \ln (5) = 0$

Add $\ln (5)$ to both sides of the equation.

$x \ln (5) = \ln (5)$

Divide each term in $x \ln (5) = \ln (5)$ by $\ln (5)$ and simplify.

Divide each term in $x \ln (5) = \ln (5)$ by $\ln (5)$ .

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (5)}{\ln (5)}$

Simplify the left side.

Cancel the common factor of $\ln (5)$ .

Cancel the common factor.

$\frac{x \ln (5)}{\ln (5)} = \frac{\ln (5)}{\ln (5)}$

Divide $x$ by $1$ .

$x = \frac{\ln (5)}{\ln (5)}$

Simplify the right side.

Cancel the common factor of $\ln (5)$ .

Cancel the common factor.

$x = \frac{\ln (5)}{\ln (5)}$

Rewrite the expression.

$x = 1$

Solve by Factoring 5^(x+1)+5^(x-1)=26

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