The given problem states that $f(x)$ is an exponential function of the form:

$f(x) = a \cdot b^x$

We are given two data points:

1. $f(3.5) = 21$
2. $f(4.5) = 46$

We need to find the value of $f(8)$ to the nearest hundredth.

Step 1: Use the data points to form equations

From the general form of the exponential function $f(x) = a \cdot b^x$, we can substitute the given values to form two equations.

For $x = 3.5$:

$f(3.5) = a \cdot b^{3.5} = 21$ $a \cdot b^{3.5} = 21 \quad \text{(Equation 1)}$

For $x = 4.5$:

$f(4.5) = a \cdot b^{4.5} = 46$ $a \cdot b^{4.5} = 46 \quad \text{(Equation 2)}$

Step 2: Solve the system of equations

We now have the system of equations:

1. $a \cdot b^{3.5} = 21$
2. $a \cdot b^{4.5} = 46$

To solve for $a$ and $b$, divide Equation 2 by Equation 1:

$\frac{a \cdot b^{4.5}}{a \cdot b^{3.5}} = \frac{46}{21}$

This simplifies to:

$b^{4.5 - 3.5} = \frac{46}{21}$ $b^1 = \frac{46}{21}$ $b = \frac{46}{21} \approx 2.1905$

Step 3: Find $a$

Substitute $b \approx 2.1905$ into Equation 1 to solve for $a$:

$a \cdot (2.1905)^{3.5} = 21$

First, calculate $(2.1905)^{3.5}$:

$(2.1905)^{3.5} \approx 17.2197$

Now solve for $a$:

$a \cdot 17.2197 = 21$ $a \approx \frac{21}{17.2197} \approx 1.2194$

Step 4: Find $f(8)$

Now that we have $a \approx 1.2194$ and $b \approx 2.1905$, we can find $f(8)$:

$f(8) = a \cdot b^8$ $f(8) = 1.2194 \cdot (2.1905)^8$

First, calculate $(2.1905)^8$:

$(2.1905)^8 \approx 221.756$

Now calculate $f(8)$:

$f(8) = 1.2194 \cdot 221.756 \approx 270.69$

Final Answer:

The value of $f(8)$ to the nearest hundredth is approximately 270.69.

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Would you like more details on any of the steps?

Here are some related questions you could explore:

1. How do you solve exponential equations generally?
2. What is the general form of an exponential function?
3. How do you calculate logarithms and use them to solve for unknowns in exponential functions?
4. What happens to the value of $f(x)$ as $x$ increases in exponential growth models?
5. How do you handle systems of equations involving exponential terms?

Tip: When solving exponential equations, it's often helpful to take the logarithm of both sides to simplify the expression.