Exponential Function Value: Calculate F(5)

When dealing with exponential functions, understanding how to predict future values based on given points is a crucial skill. These functions, often represented in the form of $y = a b^x$, describe growth or decay processes that are common in many fields, from finance to biology. In this article, we'll dive into how to find a specific value, $f(5)$, of an exponential function when we're given two other points: $f(3) = 5$ and $f(4.5) = 79$. This process involves a bit of algebraic problem-solving, but by breaking it down step-by-step, you'll see how to unravel the parameters of the function and then confidently calculate any desired value. Get ready to explore the world of exponential relationships and how to make precise predictions!

Understanding the Exponential Function Form: $y = a b^x$

The general form of an exponential function, $y = a b^x$, is foundational to understanding how these relationships work. Here, '$a$' represents the initial value or the y-intercept (the value of $y$ when $x=0$), and '$b$' is the base, which dictates the rate of growth or decay. If $b > 1$, the function exhibits exponential growth, meaning it increases at an accelerating rate. If $0 < b < 1$, it demonstrates exponential decay, decreasing at a decelerating rate. The exponent '$x$' is the variable that determines how many times the base '$b$' is multiplied by itself, scaled by the initial value '$a$'. For instance, if $a=2$ and $b=3$, the function $y=2 imes 3^x$ would start at $y=2$ (when $x=0$), then $y=6$ (when $x=1$), $y=18$ (when $x=2$), and so on. The power of this form lies in its ability to model situations where the rate of change is proportional to the current value, which is a hallmark of many real-world phenomena like compound interest, population growth, or radioactive decay. By accurately determining the values of '$a$' and '$b$', we can precisely describe and predict the behavior of such processes over time. Our task in this problem is to find these constants using the given data points and then use them to find a future value. It's like solving a puzzle where each piece of information helps us define the underlying mathematical rule.

Using Given Points to Solve for 'a' and 'b'

To determine the specific exponential function that fits our needs, we must first find the values of '$a$' and '$b$' using the information provided: $f(3) = 5$ and $f(4.5) = 79$. Since $f(x) = a b^x$, we can substitute these points into the equation to create a system of two equations:

1. For $f(3) = 5$: $5 = a b^3$
2. For $f(4.5) = 79$: $79 = a b^{4.5}$

Our goal is to solve this system for '$a$' and '$b$'. A common and effective strategy here is to divide one equation by the other. This allows us to eliminate '$a$' and solve for '$b$' first. Let's divide the second equation by the first equation:

rac{79}{5} = rac{a b^{4.5}}{a b^3}

Simplifying this gives us:

$15.8 = b^{4.5 - 3}$

$15.8 = b^{1.5}$

To find '$b$', we need to raise both sides of the equation to the power of rac{1}{1.5} (or rac{2}{3}):

b = (15.8)^{rac{1}{1.5}} = (15.8)^{rac{2}{3}}

Using a calculator, we find that $b approx 6.7225$. Now that we have the value of '$b$', we can substitute it back into either of the original equations to solve for '$a$'. Let's use the first equation, $5 = a b^3$:

$5 = a (6.7225)^3$

$5 = a (303.717...)$

Now, we solve for '$a$':

a = rac{5}{303.717...}

$a approx 0.01646$

So, our specific exponential function is approximately $f(x) = 0.01646 imes (6.7225)^x$. This process of substitution and division is key to isolating the unknown parameters of any exponential relationship when provided with sufficient data points. It's a robust method that can be applied to various similar problems, highlighting the power of algebraic manipulation in uncovering the underlying mathematical models.

Calculating $f(5)$ to the Nearest Tenth

Now that we have determined the values of '$a$' and '$b$' for our exponential function, $f(x) = a b^x$, we can calculate $f(5)$. We found that $a approx 0.01646$ and $b approx 6.7225$. So, our function is approximately $f(x) = 0.01646 imes (6.7225)^x$. To find $f(5)$, we simply substitute $x=5$ into this equation:

$f(5) = 0.01646 imes (6.7225)^5$

First, let's calculate $(6.7225)^5$:

$(6.7225)^5 approx 13541.13$

Now, multiply this by the value of '$a$':

$f(5) = 0.01646 imes 13541.13$

$f(5) approx 222.97$

The problem asks for the value of $f(5)$ to the nearest tenth. Rounding $222.97$ to the nearest tenth gives us $223.0$.

Therefore, the value of $f(5)$ for the given exponential function is approximately $223.0$. This calculation demonstrates how, once the parameters of an exponential function are established, predicting future values becomes a straightforward substitution and computation. It's a powerful tool for forecasting trends and understanding the behavior of systems that follow exponential patterns. The precision of our calculation hinges on the accuracy of the calculated '$a$' and '$b$' values, and by carrying sufficient decimal places throughout the intermediate steps, we ensure a more accurate final result. This method is widely applicable in various analytical scenarios where understanding growth or decay over time is essential.

Conclusion: The Power of Exponential Prediction

In this exploration of exponential functions, we've successfully navigated the process of determining a specific value, $f(5)$, given two points on the function's curve: $f(3) = 5$ and $f(4.5) = 79$. By leveraging the general form $y = a b^x$, we set up a system of equations, skillfully solved for the base '$b$' and the initial value '$a$', and then used these parameters to accurately calculate $f(5)$. The final result, approximately $223.0$ to the nearest tenth, showcases the predictive power inherent in understanding and applying exponential models. Whether modeling financial investments, population dynamics, or the spread of information, the ability to calculate future values from known data points is invaluable. Remember, the key lies in accurately solving for '$a$' and '$b$', which allows for precise predictions of the function's behavior at any point.

For further insights into exponential functions and their applications, you might find the resources at Khan Academy's mathematics section very helpful. They offer a comprehensive look at various mathematical concepts, including detailed explanations and practice problems on exponential growth and decay.