No Solution: Solving -2*(4/5)^(2-x)=6

Hey there, math enthusiasts! Today, we're diving into an exciting problem involving exponential equations. We'll be tackling the equation $-2 \cdot \left(\frac{4}{5}\right)^{2-x}=6$ and exploring the steps to find the solution for x. This problem combines concepts of exponents, logarithms, and algebraic manipulation, so let's break it down piece by piece to ensure we understand every step. Our goal is not just to find the answer, but to truly grasp the process and the underlying principles that make it work. So, let's get started and unravel this mathematical puzzle together!

Initial Setup and Simplification

To kick things off, let's rewrite the equation in a more manageable form. Our main goal here is to isolate the exponential term, which is the part with the variable x in the exponent. Think of it like peeling back the layers of an onion; we need to get to the core part of the equation that holds the key to our solution. To begin this isolation process, we need to get rid of that -2 that's hanging out in front of our exponential term. How do we do that? Simple – we'll divide both sides of the equation by -2. This is a fundamental algebraic principle: whatever you do to one side of the equation, you must do to the other to keep the equation balanced. So, let's perform this division and see where it takes us.

By dividing both sides by -2, we transform our original equation, $-2 \cdot \left(\frac{4}{5}\right)^{2-x}=6$, into a simpler form. Doing so, we get $\left(\frac{4}{5}\right)^{2-x} = -3$. Now, this is where things start to get interesting, and we need to put on our thinking caps. We've managed to isolate the exponential term, but we're faced with a rather peculiar situation. The right side of the equation is a negative number, -3, while the left side is an exponential expression. This is a crucial point because exponential expressions with a positive base (in our case, 4/5) will always result in a positive value, no matter what the exponent is. Think about it: raising a positive number to any power, whether it's a positive, negative, or even a fractional exponent, will never give you a negative result. This is a fundamental property of exponential functions, and it's something we need to remember.

The Implication of a Negative Result

So, what does this negative result tell us? It tells us that there's a fundamental issue with our equation. The exponential term, $\left(\frac{4}{5}\right)^{2-x}$, can never be negative. This is because no matter what value we plug in for x, raising a positive fraction (4/5) to any power will always yield a positive number. Therefore, the equation $\left(\frac{4}{5}\right)^{2-x} = -3$ has no solution. This is a critical observation and a common scenario in mathematical problem-solving. Sometimes, the equations we're given simply don't have a solution within the realm of real numbers. This can happen due to various reasons, such as conflicting conditions or, as in our case, the inherent properties of mathematical functions. Recognizing these situations is a key skill in mathematics, and it saves us from wasting time trying to find a solution that doesn't exist.

Analyzing the Proposed Solutions

Now that we've established that the equation $-2 \cdot \left(\frac{4}{5}\right)^{2-x}=6$ has no solution due to the inherent contradiction of an exponential term equaling a negative number, let's take a closer look at the proposed solutions provided in the original problem statement. These solutions offer different values for x and involve logarithmic expressions. By examining these proposed solutions, we can further solidify our understanding of why they don't work in the context of our original equation.

Examining $x=2-\log _{\frac{4}{5}}\left(\frac{1}{3}\right)$

The first proposed solution is $x=2-\log _{\frac{4}{5}}\left(\frac{1}{3}\right)$. This solution involves a logarithm with a base of 4/5 and an argument of 1/3. At first glance, this might seem like a valid solution. After all, logarithms are often used to solve exponential equations. However, we need to remember the context of our original problem. We've already determined that the equation has no solution because the exponential term can never equal a negative number. Therefore, even if this logarithmic expression yields a real number for x, it won't satisfy the original equation. To understand why, let's consider what this solution would imply if we were to substitute it back into the original equation.

If we were to plug this value of x back into the equation $\left(\frac{4}{5}\right)^{2-x} = -3$, we would be essentially saying that there exists a power to which we can raise 4/5 to get -3. But as we've already discussed, this is impossible. Raising a positive number (like 4/5) to any power will always result in a positive number. Therefore, this proposed solution, while mathematically valid in isolation, doesn't fit the constraints of our specific equation. This highlights an important aspect of problem-solving: always consider the context of the problem and whether the solutions you obtain make sense within that context. Math isn't just about crunching numbers; it's about understanding the meaning behind those numbers.

Analyzing $x=2-\log _{\frac{4}{5}}(-3)$

The second proposed solution is $x=2-\log _{\frac{4}{5}}(-3)$. This solution introduces a new issue that's even more fundamental than the one we encountered with the first proposed solution. Here, we have a logarithm with a base of 4/5, but the argument (the number inside the logarithm) is -3. This is a big red flag! Logarithms are only defined for positive arguments. The logarithm of a negative number is undefined in the realm of real numbers. This is because the logarithm function asks the question,