Solving For X In 5^(2x) + 1 = 26(5^(x-1)) A Step By Step Guide

Jul 10, 2025 by qnaunigon 63 views

Solving for x in the Equation 5^(2x) + 1 = 26(5^(x-1))

Introduction

In this article, we will delve into solving an exponential equation to find the value of x. Specifically, we will tackle the equation 5^(2x) + 1 = 26(5^(x-1)). This type of equation requires careful manipulation and a solid understanding of exponential properties to arrive at the solution. We'll break down each step in detail, making it easy to follow and comprehend. By the end of this discussion, you'll not only understand how to solve this particular problem but also gain valuable insights into handling similar exponential equations.

Understanding Exponential Equations

Before we jump into the solution, let's quickly review what exponential equations are and the basic principles involved in solving them. An exponential equation is an equation in which the variable appears in one or more exponents. These equations often require us to use properties of exponents and sometimes logarithmic functions to find the solution.

When solving exponential equations, one common strategy is to try to express all terms with the same base. This allows us to equate the exponents and solve for the variable. Another frequently used technique is substitution, which can simplify the equation by replacing a complex exponential term with a single variable. We'll be using both these techniques in solving our equation.

Keywords: exponential equations, solving equations, properties of exponents, substitution method, same base

Rewriting the Equation

Our initial equation is 5^(2x) + 1 = 26(5^(x-1)). The first step in solving this equation is to rewrite it in a more manageable form. Notice that 5^(2x) can also be written as (5^x)2. This transformation is crucial because it allows us to see the quadratic nature of the equation. Furthermore, we can simplify 5^(x-1) using the properties of exponents. Recall that a^(m-n) = a^m / a^n. Applying this, we can rewrite 5^(x-1) as 5^x / 5.

Substituting these transformations into our original equation, we get:

(5^x)2 + 1 = 26 * (5^x / 5)

Now, let's simplify the right side of the equation. 26 * (5^x / 5) can be simplified to (26/5) * 5^x. Our equation now looks like this:

(5^x)2 + 1 = (26/5) * 5^x

This form is much easier to work with and sets the stage for the next step, which involves making a strategic substitution to further simplify the equation.

Keywords: Rewriting equations, properties of exponents, simplification, (5^x)2, 5^(x-1)

Applying Substitution

To further simplify our equation (5^x)2 + 1 = (26/5) * 5^x, we'll use the substitution method. This technique is particularly useful when dealing with equations that have repeated exponential terms. Let's substitute y = 5^x. This substitution transforms the exponential terms into a simple algebraic variable, making the equation easier to handle.

Replacing 5^x with y, our equation becomes:

y^2 + 1 = (26/5) * y

This is now a quadratic equation in terms of y. To solve it, we need to rearrange it into the standard quadratic form, which is ay^2 + by + c = 0. Multiplying both sides of the equation by 5 to eliminate the fraction, we get:

5(y^2 + 1) = 26y

Expanding and rearranging the terms, we have:

5y^2 + 5 = 26y

Subtracting 26y from both sides, we get the standard quadratic form:

5y^2 - 26y + 5 = 0

Now we have a standard quadratic equation that we can solve using factoring, the quadratic formula, or completing the square. In the next section, we'll solve this quadratic equation to find the values of y.

Keywords: Substitution method, quadratic equation, standard form, y = 5^x, 5y^2 - 26y + 5 = 0

Solving the Quadratic Equation

We have arrived at the quadratic equation 5y^2 - 26y + 5 = 0. To solve this equation, we can use factoring. Factoring involves finding two binomials that multiply together to give us the quadratic expression. In this case, we are looking for two binomials that, when multiplied, result in 5y^2 - 26y + 5.

By carefully considering the coefficients, we can factor the quadratic equation as follows:

(5y - 1)(y - 5) = 0

Now, according to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. This means either (5y - 1) = 0 or (y - 5) = 0.

Solving for y in each case:

5y - 1 = 0
- Adding 1 to both sides gives 5y = 1.
- Dividing both sides by 5 gives y = 1/5.
y - 5 = 0
- Adding 5 to both sides gives y = 5.

So, we have found two possible values for y: y = 1/5 and y = 5. Now, we need to substitute back to find the corresponding values of x.

Keywords: Quadratic Equation, Factoring, Zero-product property, (5y - 1)(y - 5) = 0, y = 1/5, y = 5

Back-Substitution to Find x

In the previous section, we found the values of y for the quadratic equation. Recall that we made the substitution y = 5^x. Now, we need to substitute back these values of y to find the corresponding values of x. We have two cases to consider: y = 1/5 and y = 5.

Case 1: y = 1/5

We have 5^x = 1/5. We can rewrite 1/5 as 5^(-1). So, the equation becomes:

5^x = 5^(-1)

Since the bases are the same, we can equate the exponents:

x = -1

Case 2: y = 5

We have 5^x = 5. We can rewrite 5 as 5^1. So, the equation becomes:

5^x = 5^1

Again, since the bases are the same, we can equate the exponents:

x = 1

Therefore, we have found two solutions for x: x = -1 and x = 1. These are the values that satisfy the original equation.

Keywords: Back-substitution, y = 5^x, 5^x = 1/5, 5^x = 5, x = -1, x = 1

Verification of Solutions

To ensure the accuracy of our solutions, it's essential to verify them by substituting them back into the original equation. Our original equation was 5^(2x) + 1 = 26(5^(x-1)). We found two solutions: x = -1 and x = 1. Let's verify each one.

Verification for x = -1

Substituting x = -1 into the original equation:

5^(2*(-1)) + 1 = 26(5^(-1-1))

5^(-2) + 1 = 26(5^(-2))

1/25 + 1 = 26(1/25)

(1 + 25)/25 = 26/25

26/25 = 26/25

The equation holds true for x = -1.

Verification for x = 1

Substituting x = 1 into the original equation:

5^(2*1) + 1 = 26(5^(1-1))

5^2 + 1 = 26(5^0)

25 + 1 = 26(1)

26 = 26

The equation also holds true for x = 1.

Since both solutions satisfy the original equation, we can confidently conclude that our solutions are correct.

Keywords: Verification, substituting solutions, 5^(2x) + 1 = 26(5^(x-1)), x = -1, x = 1

Conclusion

In this article, we successfully solved the exponential equation 5^(2x) + 1 = 26(5^(x-1)). We began by rewriting the equation using the properties of exponents and then employed the substitution method to transform it into a quadratic equation. We solved the quadratic equation by factoring and found two possible values for the substituted variable. Finally, we back-substituted to find the values of x and verified our solutions by plugging them back into the original equation.

The solutions we found are x = -1 and x = 1. This exercise demonstrates the importance of understanding exponential properties and algebraic manipulation techniques in solving complex equations. By following a systematic approach and carefully applying the relevant principles, we can effectively tackle a wide range of exponential equations.

This step-by-step guide should provide you with a clear understanding of how to solve similar exponential equations in the future. Remember to practice these techniques to further enhance your problem-solving skills.

Keywords: Exponential Equation, Solving Equations, Substitution Method, Factoring, Verification, x = -1, x = 1, Mathematical Problem Solving