Solving Exponential Equations A Comprehensive Guide

Aug 10, 2025 by ADMIN 52 views

Solving Tricky Exponential Equations: A Step-by-Step Guide

Hey guys! Let's dive into some interesting math problems involving exponential equations. These types of questions might seem daunting at first, but with a systematic approach, they become much easier to handle. We're going to break down three different problems, showing you the tricks and techniques to solve them. So, grab your calculators, and let's get started!

1. (x² - 10x + 24)^(2x-3) = (x² - 10x + 24)^(x+1)

This is an interesting equation where we have the same base raised to different powers. To solve this, we need to consider the scenarios where the equation holds true. Our main keyword here is solving exponential equations with the same base. The key idea is that if a^m = a^n, then either m = n or the base a has specific values that make the equation true regardless of the exponents. Let's break it down step by step.

First, let's consider the case where the exponents are equal. We have 2x - 3 = x + 1. Solving this linear equation, we get:

2x - 3 = x + 1
2x - x = 1 + 3
x = 4

So, x = 4 is one solution. But we can't stop here! We need to consider other possibilities. Remember, if the base is 1, the equation holds true regardless of the exponent. So, let's set the base equal to 1:

x² - 10x + 24 = 1
x² - 10x + 23 = 0

Now we have a quadratic equation. To solve it, we can use the quadratic formula:

x = [-b ± √(b² - 4ac)] / 2a

In our case, a = 1, b = -10, and c = 23. Plugging these values in, we get:

x = [10 ± √((-10)² - 4 * 1 * 23)] / 2 * 1
x = [10 ± √(100 - 92)] / 2
x = [10 ± √8] / 2
x = [10 ± 2√2] / 2
x = 5 ± √2

So, we have two more potential solutions: x = 5 + √2 and x = 5 - √2. But wait, there's more! We also need to consider the case where the base is 0. If the base is 0, the exponents must be positive for the equation to hold. Let's set the base equal to 0:

x² - 10x + 24 = 0
(x - 6)(x - 4) = 0

This gives us x = 6 and x = 4. We already found x = 4 as a solution, but we need to check if x = 6 works. Let's plug x = 6 into the exponents:

2x - 3 = 2(6) - 3 = 9 (positive)
x + 1 = 6 + 1 = 7 (positive)

Since both exponents are positive when x = 6, this is also a valid solution. Finally, we need to consider the case where the base is -1. If the base is -1, the exponents must both be even or both be odd for the equation to hold. Let's set the base equal to -1:

x² - 10x + 24 = -1
x² - 10x + 25 = 0
(x - 5)² = 0
x = 5

Now, let's plug x = 5 into the exponents:

2x - 3 = 2(5) - 3 = 7 (odd)
x + 1 = 5 + 1 = 6 (even)

Since one exponent is odd and the other is even, x = 5 is not a solution. So, the solutions to the equation are x = 4, x = 6, x = 5 + √2, and x = 5 - √2. That was quite a journey, but we nailed it!

Key Takeaways for Solving Equations with the Same Base

Equate the exponents: If a^m = a^n, then m = n. This is the most straightforward case.
Consider the base equals 1: If the base is 1, the equation holds true regardless of the exponents.
Consider the base equals 0: If the base is 0, both exponents must be positive.
Consider the base equals -1: If the base is -1, both exponents must be even or both must be odd.

2. 5^(2x) - 30 * 5^x + 125 = 0

This equation looks a bit different, but we can use a clever substitution to make it easier to solve. Our main keyword for this section is solving exponential equations using substitution. The trick here is to recognize that 5^(2x) can be written as (5^x)². This allows us to transform the equation into a quadratic form. Let's see how it works.

Let's substitute y = 5^x. Then, the equation becomes:

y² - 30y + 125 = 0

Now we have a quadratic equation in terms of y. We can solve this by factoring:

(y - 25)(y - 5) = 0

This gives us two possible values for y: y = 25 and y = 5. But remember, we're not trying to solve for y; we want to solve for x. So, we need to substitute back 5^x for y.

For y = 25, we have:

5^x = 25
5^x = 5²
x = 2

For y = 5, we have:

5^x = 5
5^x = 5¹
x = 1

So, the solutions to the equation are x = 2 and x = 1. See how substitution made the problem much more manageable? By recognizing the structure of the equation, we could transform it into a familiar form and solve it easily.

Tips for Using Substitution in Exponential Equations

Identify the repeating exponential term: Look for terms like a^(2x), which can be written as (a^x)².
Make the substitution: Let y = a^x or any other appropriate variable.
Solve the resulting equation: This will usually be a quadratic equation.
Substitute back: Replace the variable with the original exponential term.
Solve for x: Use logarithms or other techniques to find the value of x.

3. 4^x + 2^(5-2x) = 12

This equation looks tricky, but we can manipulate it to make it solvable. Our keyword here is manipulating exponential equations for simpler solving. The key is to express all terms with the same base. Notice that 4 is a power of 2 (4 = 2²). Let's use this to rewrite the equation.

We can rewrite 4^x as (2²)^x, which is the same as 2^(2x). Also, we can rewrite 2^(5-2x) as 2^5 / 2^(2x). So, the equation becomes:

2^(2x) + 2^5 / 2^(2x) = 12

Now, let's substitute y = 2^(2x). The equation becomes:

y + 32 / y = 12

To get rid of the fraction, we can multiply both sides by y:

y² + 32 = 12y
y² - 12y + 32 = 0

Now we have a quadratic equation in terms of y. We can solve this by factoring:

(y - 8)(y - 4) = 0

This gives us two possible values for y: y = 8 and y = 4. Now, we need to substitute back 2^(2x) for y.

For y = 8, we have:

2^(2x) = 8
2^(2x) = 2³
2x = 3
x = 3/2

For y = 4, we have:

2^(2x) = 4
2^(2x) = 2²
2x = 2
x = 1

So, the solutions to the equation are x = 3/2 and x = 1. We successfully solved another challenging exponential equation by manipulating the terms and using substitution!

Strategies for Manipulating Exponential Equations

Express all terms with the same base: This is often the first step to simplifying the equation.
Use exponent rules: Remember rules like (a^m)n = a^(mn) and a^(m+n) = a^m * a^n.
Look for opportunities to substitute: If you see repeating exponential terms, substitution can simplify the equation.
Multiply to eliminate fractions: If there are fractions with exponential terms in the denominator, multiplying through can help.

Conclusion

So, there you have it, guys! We've tackled three different types of exponential equations and learned some powerful techniques for solving them. Remember, the key is to break down the problem into smaller, manageable steps. By considering different cases, using substitution, and manipulating the equation, you can conquer even the trickiest exponential problems. Keep practicing, and you'll become a master of exponential equations in no time!

I hope this guide has been helpful. If you have any questions or want to try more examples, feel free to ask. Happy solving!