Solve Exponential Equations: Step-by-Step Guide

Hey guys! Today, we're diving deep into the fascinating world of exponential equations. These equations, where the variable lurks in the exponent, might seem intimidating at first glance, but trust me, with the right techniques, they become a piece of cake. We'll tackle two specific problems, breaking down each step and revealing the underlying principles. So, buckle up and let's get started!

Solving Exponential Equations: A Step-by-Step Approach

i) Deciphering 2⁻ˣ = 8 [3 Marks]

This first equation, 2⁻ˣ = 8, is a classic example of how we can manipulate exponents to find our solution. The key idea here is to express both sides of the equation with the same base. Why? Because if we have the same base, we can simply equate the exponents.

Let's break it down:

1. Expressing 8 as a power of 2: We know that 8 can be written as 2³, which means 2 x 2 x 2 = 8. This is a fundamental step in solving many exponential equations, so it's crucial to recognize these common powers quickly.
2. Rewriting the equation: Now we can rewrite our original equation, 2⁻ˣ = 8, as 2⁻ˣ = 2³. See how both sides now have the same base (2)? This is exactly what we wanted!
3. Equating the exponents: Since the bases are the same, we can confidently equate the exponents. This gives us the simple linear equation: -x = 3.
4. Solving for x: To isolate x, we multiply both sides of the equation by -1. This gives us our solution: x = -3.

Therefore, the solution to the equation 2⁻ˣ = 8 is x = -3.

Key Takeaway: The core strategy here is to manipulate the equation so that both sides have the same base. Once you achieve this, the problem reduces to solving a much simpler equation involving the exponents. This technique is widely applicable to a variety of exponential equations, making it a crucial tool in your mathematical arsenal. Remember to always look for ways to express numbers as powers of a common base, and you'll be well on your way to mastering exponential equations!

ii) Taming 5ˣ + 125(5⁻ˣ) = 30 [6 Marks]

Now, let's tackle the second equation: 5ˣ + 125(5⁻ˣ) = 30. This one looks a bit more complex, but don't worry, we'll break it down step by step. The trick here involves a clever substitution that transforms the equation into a more familiar quadratic form. This is a common technique used to simplify equations that might initially seem daunting.

1. Dealing with the negative exponent: The term 5⁻ˣ can be a bit tricky to work with directly. Remember that a negative exponent means we take the reciprocal. So, 5⁻ˣ is the same as 1/5ˣ. This is a fundamental property of exponents that is essential for simplifying expressions.
2. Rewriting the equation: Substituting 1/5ˣ for 5⁻ˣ, our equation becomes: 5ˣ + 125(1/5ˣ) = 30. This might look slightly better, but we still have that pesky fraction to deal with.
3. Eliminating the fraction: To get rid of the fraction, we can multiply the entire equation by 5ˣ. This gives us: (5ˣ)² + 125 = 30(5ˣ). Multiplying each term ensures that the equation remains balanced and the solution set remains the same.
4. Recognizing the quadratic form: Now, let's rearrange the equation: (5ˣ)² - 30(5ˣ) + 125 = 0. Do you see it? This equation has the form of a quadratic equation! It's like ax² + bx + c = 0, where our variable is not x, but rather 5ˣ. This recognition is a crucial step in solving the problem.
5. Introducing a substitution: To make things even clearer, let's make a substitution. Let y = 5ˣ. This simplifies our equation to: y² - 30y + 125 = 0. Now it's undeniably a quadratic equation, and we have familiar tools to solve it.
6. Solving the quadratic equation: We can solve this quadratic equation by factoring. We need to find two numbers that multiply to 125 and add up to -30. These numbers are -5 and -25. So, we can factor the equation as: (y - 5)(y - 25) = 0.
7. Finding the values of y: For the product of two factors to be zero, at least one of them must be zero. Therefore, we have two possible solutions for y: y = 5 or y = 25.
8. Substituting back for x: Remember that y = 5ˣ. Now we need to substitute back to find the values of x.
   • If y = 5, then 5ˣ = 5. Since 5¹ = 5, we have x = 1.
   • If y = 25, then 5ˣ = 25. Since 5² = 25, we have x = 2.

Therefore, the solutions to the equation 5ˣ + 125(5⁻ˣ) = 30 are x = 1 and x = 2.

Key Takeaways: This problem highlights the power of substitution in simplifying complex equations. By recognizing the quadratic form and making the appropriate substitution, we transformed a seemingly difficult exponential equation into a manageable quadratic. Remember to always look for patterns and opportunities to simplify equations, and don't be afraid to introduce new variables to make the problem clearer. This technique is not just limited to exponential equations; it's a valuable problem-solving skill in many areas of mathematics.

Mastering Exponential Equations: Practice Makes Perfect

Exponential equations are a fundamental concept in mathematics, appearing in various fields like calculus, physics, and finance. Mastering these equations is not just about memorizing steps; it's about understanding the underlying principles and developing a strategic approach to problem-solving. The more you practice, the more comfortable you'll become with recognizing patterns, applying the right techniques, and finding those elusive solutions. So, keep practicing, keep exploring, and keep unlocking the power of exponential equations!