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

Solution:

Step 1:

Eliminate $26$ from each side of the equation to get $5^{x + 1} + 5^{x - 1} - 26 = 0$.

Step 2:

Extract $5^{x - 1}$ as a common factor to rewrite the equation as $5^{x - 1}(5^2 + 1)$.

Step 3:

Reintroduce $26$ to the equation, resulting in $5^{x + 1} + 5^{x - 1} = 26$.

Step 4:

Combine $25$ and $1$ within the factored term to obtain $5^{x - 1} \cdot 26$.

Step 5:

Rearrange to place $26$ before $5^{x - 1}$, yielding $26 \cdot 5^{x - 1} = 26$.

Step 6:

Divide the equation $26 \cdot 5^{x - 1} = 26$ by $26$ to isolate $5^{x - 1}$.

Step 6.1:

Perform the division $\frac{26 \cdot 5^{x - 1}}{26} = \frac{26}{26}$.

Step 6.2:

Simplify the left side by canceling out $26$.

Step 6.2.1:

Eliminate the common factor of $26$ to get $\frac{\cancel{26} \cdot 5^{x - 1}}{\cancel{26}} = \frac{26}{26}$.

Step 6.2.1.1:

Complete the division to find $5^{x - 1} = \frac{26}{26}$.

Step 6.3:

Simplify the right side to $5^{x - 1} = 1$.

Step 7:

Apply the natural logarithm to both sides, $\ln(5^{x - 1}) = \ln(1)$, to solve for $x$.

Step 8:

Use the logarithmic property to bring down the exponent: $(x - 1) \ln(5) = \ln(1)$.

Step 9:

Distribute the logarithm on the left side.

Step 9.1:

Apply the distributive property to get $x \ln(5) - \ln(5) = \ln(1)$.

Step 9.1.1:

Simplify the expression to $x \ln(5) - \ln(5) = \ln(1)$.

Step 10:

Recognize that $\ln(1)$ is $0$ and simplify to $x \ln(5) - \ln(5) = 0$.

Step 11:

Add $\ln(5)$ to both sides to isolate the term with $x$: $x \ln(5) = \ln(5)$.

Step 12:

Divide by $\ln(5)$ to solve for $x$.

Step 12.1:

Divide each term to get $\frac{x \ln(5)}{\ln(5)} = \frac{\ln(5)}{\ln(5)}$.

Step 12.2:

Simplify the left side by canceling out $\ln(5)$.

Step 12.2.1:

Cancel the common factor to find $x = \frac{\ln(5)}{\ln(5)}$.

Step 12.3:

Simplify the right side to conclude that $x = 1$.

Knowledge Notes:

The problem-solving process involves several mathematical concepts and properties:

1. Exponentiation: The original equation involves $5^{x+1}$ and $5^{x-1}$, which are exponential expressions where $5$ is the base and $x+1$ and $x-1$ are the exponents.

2. Factoring: Factoring is a process of breaking down an expression into a product of simpler expressions. In this case, $5^{x-1}$ is factored out.

3. Logarithms: The natural logarithm, denoted as $\ln$, is the inverse operation of exponentiation with base $e$. It is used to solve for variables in the exponent.

4. Logarithmic Properties: The property $\ln(a^b) = b \ln(a)$ is used to bring down the exponent in front of the logarithm, which is a key step in solving the equation.

5. Simplification: The process involves simplifying expressions by canceling common factors and recognizing that $\ln(1) = 0$.

6. Isolation of Variables: The goal is to isolate the variable $x$ on one side of the equation to solve for it. This involves adding, subtracting, multiplying, or dividing both sides of the equation by the same value.

Understanding these concepts is essential for solving exponential equations and manipulating expressions involving logarithms.