Rewrite Exponential Functions: The $f(t)=ab^t$ Form

Hey guys, let's dive into the awesome world of exponential functions! You know, those functions where the variable is chilling in the exponent? Today, we're going to tackle a specific problem: taking an exponential function that looks a little bit tricky and rewriting it into a simpler, more standard form. The goal is to get our function, which is currently $f(t)=900(1.14)^{3 t}$, into the super common $f(t)=a b^t$ format. We'll also make sure all our numbers (coefficients) are rounded super neatly to four decimal places. So, buckle up, grab your calculators, and let's get this mathematical party started! Understanding how to manipulate these functions is key to seeing their underlying growth or decay patterns more clearly.

Why the $f(t)=ab^t$ Form is Your New Best Friend

So, why bother rewriting our function into the $f(t)=a b^t$ form, you ask? Well, think of it like this: this form is the universal language of exponential growth and decay. When a function is in this $f(t)=a b^t$ shape, it's super easy to spot the key players. The 'a' value is your initial amount – what you start with at time $t=0$. The 'b' value is your growth or decay factor. If 'b' is greater than 1, your function is growing; if 'b' is between 0 and 1, it's shrinking. And the 't' is just your time variable. Seeing these components laid out so clearly makes it a breeze to understand the behavior of the function. For our specific problem, $f(t)=900(1.14)^{3 t}$, it's already got the initial amount $900$ staring us in the face. However, the exponent, $3t$, is a bit clunky. It means the growth factor $1.14$ is being applied three times per unit of time. By converting it to $f(t)=a b^t$, we're essentially finding a new, single growth factor 'b' that represents the effective growth over one unit of time, simplifying the analysis. This standardization is crucial in many fields, from finance (compound interest) to biology (population growth) and physics (radioactive decay). It allows for direct comparison between different exponential processes.

Let's Get Down to Business: Rewriting the Function

Alright, team, let's roll up our sleeves and do the math! Our starting point is $f(t)=900(1.14)^{3 t}$. We want to transform this into the $f(t)=a b^t$ format. First off, notice that the $900$ is already in the right place for our 'a' value. It's the initial amount. So, we know that a = 900. Now, the tricky part is dealing with that $(1.14)^{3 t}$ term. Using the magic of exponent rules, specifically $(x^m)^n = x^{mn}$, we can rewrite $(1.14)^{3 t}$ as $((1.14)^3)^t$. This is super important because now we have 't' as the only exponent, which is exactly what we need for the $f(t)=a b^t$ form! Our function now looks like $f(t)=900((1.14)^3)^t$. The base of our new exponential term, $((1.14)^3)$, is going to be our new 'b' value. So, b = (1.14)^3. Now, we just need to calculate this value and round it to four decimal places.

Let's crunch those numbers: $(1.14)^3 = 1.14 * 1.14 * 1.14 = 1.2996 * 1.14 = 1.481544$.

We need to round this to four decimal places. The fifth decimal place is a 4, so we round down. This gives us b ≈ 1.4815. Therefore, our function rewritten in the desired form is $f(t) = 900(1.4815)^t$. See? We took a function that looked a little complicated and simplified it into a clear, understandable form. This new form clearly shows us that the initial amount is $900$ and the effective growth factor per unit of time is approximately $1.4815$, meaning the quantity increases by about $48.15$% each time period. This transformation is incredibly useful for comparing growth rates directly. If you had another function like $g(t)=900(1.2)^t$, you could immediately see that $f(t)$ grows faster because $1.4815 > 1.2$. This kind of insight is invaluable when analyzing data or making predictions.

Understanding the Transformation: A Deeper Dive

Let's really break down why this transformation works and what it means conceptually. We started with $f(t)=900(1.14)^{3 t}$. The core idea here is that the exponent $3t$ means the base $1.14$ is being multiplied by itself $3t$ times. Now, think about the properties of exponents. One fundamental rule is $(x^m)^n = x^{m imes n}$. We can use this rule in reverse, or rather, we can use it to rearrange our expression. We want our exponent to be just 't', so we need to isolate it. We can rewrite $3t$ as $3 imes t$. So, our expression becomes $f(t) = 900(1.14)^{3 imes t}$. Now, applying that exponent rule in reverse, we can see that $(1.14)^{3 imes t}$ is the same as $((1.14)^3)^t$. This is the key step! We've essentially bundled the repeated multiplication of $1.14$ into a single, new base. Instead of multiplying by $1.14$ three times for every unit increase in $t$, we're now multiplying by a single number, $(1.14)^3$, just once for every unit increase in $t$. This new base, $(1.14)^3$, represents the effective growth factor over one unit of time. When we calculate $(1.14)^3$, we get approximately $1.481544$. Rounding this to four decimal places gives us $1.4815$. So, the function $f(t)=900(1.14)^{3 t}$ is mathematically equivalent to $f(t)=900(1.4815)^t$. The initial value $a=900$ remains unchanged because when $t=0$, $(1.14)^{3 imes 0} = (1.14)^0 = 1$, and similarly, $(1.4815)^0 = 1$. The change only affects the base that dictates the rate of growth or decay over time. This process is vital for comparing different exponential models side-by-side. For instance, if you're looking at two investments, one compounding interest daily and another monthly, you'd want to express their growth rates in a comparable form, often an annual growth factor, to make an informed decision. This type of algebraic manipulation allows us to do just that.

Final Answer and What It Means

So, after all that number crunching and exponent wrangling, we've arrived at our final answer! The function $f(t)=900(1.14)^{3 t}$, when rewritten in the form $f(t)=a b^t$ and with coefficients rounded to four decimal places, is $f(t) = 900(1.4815)^t$. Here, our $a$ value is $900$, representing the initial amount. Our $b$ value is approximately $1.4815$, which is our new, effective growth factor. Since $b > 1$, we know this function represents exponential growth. Specifically, the quantity is increasing by about $48.15$% per unit of time. This form is incredibly useful because it provides an immediate snapshot of the function's behavior: where it starts and how fast it's changing. Imagine you're tracking the population of a city or the value of an investment. This standardized form lets you quickly compare different scenarios. For example, if you had another city's population growth modeled as $g(t)=900(1.4)^t$, you could instantly tell that the first city ($f(t)$) is growing faster because $1.4815$ is greater than $1.4$. This is the power of standardization in mathematics – it makes complex information digestible and comparable. So, next time you see an exponential function with a slightly more complex exponent, remember you can always simplify it into this elegant $f(t)=a b^t$ form to better understand its dynamics. Keep practicing, and you'll be a master of exponential functions in no time! This skill is not just for math class; it's a fundamental tool for understanding growth and change in the real world, from economic trends to scientific discoveries.