Exponential Function Equation: Find An Equivalent Form

by Andrew McMorgan 55 views

Hey guys, let's dive into the cool world of exponential functions! Today, we've got a challenge from Pablo involving a function he cooked up: f(x)= rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}. He wants to know which representation is equivalent to this gem. This is all about understanding how exponential functions work and how we can manipulate their equations. We'll be looking at the relationship between consecutive terms in a sequence defined by an exponential function. Think of it like finding a secret code that describes how the sequence grows or shrinks. We'll break down the original function, explore the properties of exponential growth, and then rigorously test each option to see which one fits. This isn't just about memorizing formulas; it's about understanding the DNA of these functions. Get ready to flex those math muscles!

Deconstructing Pablo's Function: The Core of the Problem

Alright, let's get down to business with Pablo's function: f(x)= rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}. The key here is to recognize this as an exponential function. We have a base, which is $\\frac{5}{2}\$, and an exponent, x−1x-1. The $\\frac{3}{2}\$ is our initial value or coefficient. This function tells us the value of the xth x^{\text {th }} number in a sequence. Exponential functions are super powerful because they describe situations where something grows or decays at a rate proportional to its current value. Think of compound interest, population growth, or radioactive decay – all ruled by these functions. The general form of an exponential function is often written as f(x)=a⋅bxf(x) = a \\\cdot \\ b^{x}, where aa is the initial value and bb is the growth factor. In Pablo's case, the x−1x-1 in the exponent is a slight variation, but it's still fundamentally exponential. Understanding this base and how the exponent affects the output is crucial. The base, $\\frac{5}{2}\$, tells us that for every unit increase in xx, the value of the function is multiplied by $\\frac{5}{2}\$. This multiplicative relationship is the hallmark of exponential sequences. We're essentially looking for an equivalent way to express this multiplicative relationship between terms.

Exploring the Options: A Deep Dive into Equivalence

Now, let's scrutinize the potential equivalent representations Pablo has given us. We need to see which one perfectly mirrors the behavior of his original function, f(x)= rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}. Let's take a peek at the options:

Option A: f(x+1)= rac{5}{2} f(x)

This option suggests a relationship between the term at position x+1x+1 and the term at position xx. It implies that to get the next term in the sequence, you multiply the current term by $\\frac{5}{2}\$. Let's test this. We need to find f(x+1)f(x+1) from Pablo's original function. If f(x)= rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}, then f(x+1)f(x+1) would be obtained by replacing every xx with (x+1)(x+1):

f(x+1) = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{(x+1)-1} = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}

Now, let's look at the right side of option A: 52f(x)\\\frac{5}{2} f(x). Substituting Pablo's function for f(x)f(x), we get:

\\\frac{5}{2} \\left( rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1} \\right) = rac{5}{2} \\\cdot \\ rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}

To compare this with f(x+1)f(x+1), we need to simplify. Remember the exponent rule am⋅an=am+na^m \\\cdot \\ a^n = a^{m+n} and am−n=am/ana^{m-n} = a^m / a^n. Let's rewrite 52$(52)x−1\\\frac{5}{2}\\\$\\\left(\\\frac{5}{2}\\\right)^{x-1} using exponent properties. We can write 52$(52)x−1\\\frac{5}{2}\\\$\\\left(\\\frac{5}{2}\\\right)^{x-1} as \\\frac{5}{2}\\\$\\\left(\\\frac{5}{2}\\\right)^{x} \\cdot \\ rac{5}{2}\\\left(\\\frac{5}{2}\\\right)^{-1}. Or, more simply, we can add the exponents: 1+(x−1)=x1 + (x-1) = x. So, the expression becomes:

\\\frac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1} \\cdot \\ rac{5}{2} = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1+1} = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}

Look at that! Both f(x+1)f(x+1) and 52f(x)\\\frac{5}{2} f(x) simplify to rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}. This means option A holds true! It correctly captures the recursive relationship where each term is $\\frac{5}{2}\$\\text{ times the previous term. This is the essence of geometric sequences! The fact that the relationship holds for any xx is what makes it an equivalent representation.

Option B: f(x)= rac{5}{2} f(x+1)

Let's check out option B. This one suggests that the current term f(x)f(x) is 52$ times the *next* term, f(x+1)\\\frac{5}{2}\\\$\\\text{ times the *next* term, } f(x+1). This sounds a bit backward compared to how sequences usually grow. Let's use what we found earlier: f(x+1) = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x} and f(x) = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}.

So, the right side of option B is:

\\\frac{5}{2} f(x+1) = rac{5}{2} \\left( rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x} \\right) = rac{5}{2} \\cdot \\ rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}

To make this comparable to f(x)f(x), we can again use exponent rules. \\\frac{5}{2} imes rac{3}{2} imes rac{5}{2}^x = rac{3}{2} imes rac{5}{2} imes rac{5}{2}^x = rac{3}{2} imes rac{5}{2}^{x+1}.

Now, let's compare this to f(x) = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}. Clearly, rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x+1} is not the same as rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}. They differ significantly, especially as xx changes. Thus, option B is not an equivalent representation.

Option C: The "Discussion category : mathematics" Placeholder

This isn't an equation at all, guys! It's just a category tag. So, obviously, it can't be an equivalent mathematical representation. We can safely ignore this one for finding the right equation. It's important to focus on the mathematical expressions when we're asked for mathematical equivalence.

The Verdict: Confirming the Equivalent Representation

After our thorough analysis, we found that Option A: f(x+1)= rac{5}{2} f(x) is the only one that accurately reflects the relationship inherent in Pablo's original exponential function. This equation tells us that each term in the sequence is $\\frac5}{2}\$\\text{ times the preceding term. This is the defining characteristic of a geometric sequence where the common ratio is $\\frac{5}{2}\$. The original function f(x)= rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1} can be rewritten as f(x) = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}\\\left(\\\frac{5}{2}\\\right)^{-1} = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}\\\left(\\\frac{2}{5}\\\right) = rac{3}{5}\\\left(\\\frac{5}{2}\\\right)^{x}. If we use this form, we can see the relationship more clearly. Let's check f(x+1)f(x+1) with this adjusted form $f(x+1) = rac{35}\\left(\\frac{5}{2}\\right)^{x+1}$. Now, let's check 52f(x)\\\frac{5}{2} f(x) $\\frac{5{2} \left( rac{3}{5}\\left(\\frac{5}{2}\\right)^{x} \right) = rac{5}{2}\\left(\\frac{3}{5}\\right)\\left(\\frac{5}{2}\\right)^{x} = rac{3}{2}\\left(\\frac{5}{2}\\right)^{x}$. This seems a bit off in my re-check. Let me go back to the original manipulation which was correct.

Let's re-verify the algebra for Option A using the original form of f(x) = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1}.

We want to show that f(x+1) = rac{5}{2} f(x).

Left side: f(x+1) = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{(x+1)-1} = rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}.

Right side: rac{5}{2} f(x) = rac{5}{2} \\left( rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1} \\right).

To simplify the right side, we can distribute the 52$ into the parentheses. Using the exponent rule amcdotan=am+n\\\frac{5}{2}\\\$\\\text{ into the parentheses. Using the exponent rule } a^m \\cdot a^n = a^{m+n}, we have:

\\\frac{5}{2} \\cdot rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x-1} = rac{3}{2} \\cdot \\left( rac{5}{2} \\cdot rac{5}{2}^{x-1} \\right) = rac{3}{2} \\cdot rac{5}{2}^{1 + (x-1)} = rac{3}{2} \\cdot rac{5}{2}^{x}.

So, the left side f(x+1)f(x+1) equals rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}, and the right side rac{5}{2} f(x) also equals rac{3}{2}\\\left(\\\frac{5}{2}\\\right)^{x}. Therefore, f(x+1)= rac{5}{2} f(x) is indeed the correct equivalent representation. This confirms that the sequence generated by f(x)f(x) is a geometric sequence with a common ratio of $\\frac{5}{2}\$.

Conclusion: Understanding the Pattern

In essence, Pablo's function describes a sequence where each term is \\\frac{5}{2}\\\$\\\text{ times the previous term. Option A, } f(x+1)= rac{5}{2} f(x), is the most direct and elegant way to express this fundamental property of geometric sequences. It's a powerful reminder that equivalent representations can highlight different aspects of a function, whether it's its explicit formula or its recursive relationship. Keep practicing, and you'll master these transformations in no time! Math on, everyone!