Solving Exponential Equations A Step By Step Guide

Hey everyone! Today, we're diving deep into the world of exponential equations and tackling a specific problem: how to solve the equation $10^{-x} = 6^{5x}$. Exponential equations can seem intimidating at first, but with the right approach and a few key logarithmic properties, they become much more manageable. So, grab your calculators, and let's get started!

Understanding Exponential Equations

First things first, what exactly are exponential equations? In a nutshell, they are equations where the variable appears in the exponent. The equation $10^{-x} = 6^{5x}$ perfectly exemplifies this, with 'x' playing a crucial role in the exponents of both 10 and 6. Mastering exponential equations is vital because they pop up in numerous real-world scenarios, from calculating compound interest and modeling population growth to understanding radioactive decay and analyzing chemical reactions. They're not just abstract mathematical concepts; they're powerful tools for describing and predicting change in various systems.

The key to unlocking these equations lies in understanding logarithms. Logarithms are essentially the inverse operation of exponentiation. Think of it this way: if exponentiation is raising a base to a power, logarithms are asking, "What power do I need to raise this base to, in order to get this number?" This inverse relationship is what allows us to isolate the variable in exponential equations. When you encounter an equation where the variable is stuck up in the exponent, your first instinct should be to consider using logarithms. Logarithms are the magic key that unlocks exponential equations!

There are two main types of logarithms we often encounter: common logarithms (base 10) and natural logarithms (base e, where e is approximately 2.71828). Common logarithms are denoted as "log," and natural logarithms are denoted as "ln." The choice of which logarithm to use often comes down to convenience or the specific context of the problem. In many cases, either type will work just fine, as long as you apply the logarithmic properties correctly. We will be using common logarithms in this solution.

Before we dive into the solution, let's quickly review some essential logarithmic properties that we'll be using. These properties are the bread and butter of solving exponential equations, and understanding them thoroughly will make the process much smoother.

Key Logarithmic Properties

1. The Power Rule: log_b(a^c) = c * log_b(a). This property is a game-changer! It allows us to bring exponents down as coefficients, effectively freeing the variable from its exponential prison. In our case, it will allow us to get the 'x' out of the exponents in our equation.
2. The Logarithm of a Product: log_b(xy) = log_b(x) + log_b(y). While we won't be directly using this property in this particular problem, it's a valuable tool in other scenarios where you have the logarithm of a product.
3. The Logarithm of a Quotient: log_b(x/y) = log_b(x) - log_b(y). Similar to the product rule, this property can simplify expressions involving logarithms of quotients.
4. log_b(b) = 1. The logarithm of a number to the same base is always 1. For instance, log₁₀(10) = 1 and ln(e) = 1. This property is helpful in simplifying expressions after applying other logarithmic rules.
5. log_b(1) = 0. The logarithm of 1 to any base is always 0. This is because any number raised to the power of 0 equals 1.

With these properties in our arsenal, we're well-equipped to tackle the equation $10^{-x} = 6^{5x}$. Let's move on to the step-by-step solution!

Solving the Exponential Equation $10^{-x} = 6^{5x}$

Now, let's break down the process of solving the equation $10^{-x} = 6^{5x}$ step by step. Remember, the key to solving exponential equations is to use logarithms to bring the variable down from the exponent. Here's how we'll do it:

Step 1: Take the logarithm of both sides.

To begin, we apply the logarithm function to both sides of the equation. We can use either the common logarithm (log base 10) or the natural logarithm (ln), but for this example, let's stick with the common logarithm (log). Applying the logarithm to both sides gives us:

log(10^-x) = log(6^5x)

Why do we do this? By taking the logarithm of both sides, we set the stage for using the power rule of logarithms, which will allow us to bring the exponents down as coefficients.

Step 2: Apply the power rule of logarithms.

This is where the magic happens! The power rule states that log_b(a^c) = c * log_b(a). We apply this rule to both sides of our equation:

-x * log(10) = 5x * log(6)

Notice how the exponents, -x and 5x, have now become coefficients. This is a crucial step in isolating the variable 'x'. The power rule is our best friend when dealing with exponential equations!

Step 3: Simplify the equation.

We can simplify the left side of the equation further. Remember that log(10) is the logarithm base 10 of 10, which equals 1. So, we have:

-x * 1 = 5x * log(6)

This simplifies to:

-x = 5x * log(6)

Step 4: Isolate the terms with 'x'.

Our goal is to get all the terms involving 'x' on one side of the equation. To do this, we can add 'x' to both sides:

0 = 5x * log(6) + x

Step 5: Factor out 'x'.

Now, we can factor out 'x' from the right side of the equation:

0 = x * (5log(6) + 1)

Step 6: Solve for 'x'.

We now have a product of two factors that equals zero. This means that either x = 0 or (5log(6) + 1) = 0. Let's consider each case:

• Case 1: x = 0

This is one possible solution. If x = 0, the equation holds true because $10^{-0} = 1$ and $6^{5*0} = 6^0 = 1$.

• Case 2: 5log(6) + 1 = 0

To solve for 'x' in this case, we need to isolate 'x'. This part requires some more algebraic manipulation. We start by subtracting 1 from both sides:

5log(6) = -1

Wait a minute! There's no 'x' in this equation anymore. It seems we took a wrong turn somewhere. Let's rewind back to Step 5.

In Step 5, we correctly factored out 'x', leading to the equation 0 = x * (5log(6) + 1). The error lies in assuming that the second factor can be equal to zero and trying to solve from there. Actually, we should divide both sides of the equation by (5log(6) + 1). We can do this as long as (5log(6) + 1) is not zero. Let's evaluate this expression:

5log(6) + 1 ≈ 5 * 0.778 + 1 ≈ 3.89 + 1 ≈ 4.89

Since this value is not zero, we can indeed divide both sides of the equation 0 = x * (5log(6) + 1) by (5log(6) + 1), which directly gives us x = 0. It means we had the correct solution all along!

So, there was a little bit of a detour there, but we learned a valuable lesson about carefully considering each step and making sure our algebraic manipulations are valid. Always double-check your work, guys!

Step 7: State the solution.

The only solution to the equation $10^{-x} = 6^{5x}$ is x = 0.

Verification

It's always a good idea to verify our solution by plugging it back into the original equation. Let's substitute x = 0 into the equation $10^{-x} = 6^{5x}$:

$10^{-0} = 6^{5*0}$

$10^0 = 6^0$

1 = 1

The equation holds true, so our solution x = 0 is correct. Always verify your solutions, guys. It's a simple step that can save you from making mistakes.

Conclusion

There you have it! We've successfully solved the exponential equation $10^{-x} = 6^{5x}$. Remember, the key to solving exponential equations is to use logarithms to bring the variable down from the exponent. By applying the power rule of logarithms and carefully following the steps, we were able to isolate 'x' and find the solution.

Don't be intimidated by exponential equations. With a solid understanding of logarithms and their properties, you can conquer any exponential equation that comes your way. Keep practicing, and you'll become a pro at solving these types of problems in no time! Keep up the great work, guys!

Final Answer:

A. x = 0