We are given that the function $f(x)$ follows an exponential form:

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

Step 1: Setting Up the Equations

From the given values:

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

Step 2: Finding the Ratio

Dividing the second equation by the first:

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

Since $a$ cancels out, we get:

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

$b^1 = \frac{46}{21}$

$b = \frac{46}{21} \approx 2.19$

Step 3: Solving for $a$

Substituting $b \approx 2.19$ into the first equation:

$a \cdot (2.19)^{3.5} = 21$

Taking logarithms or using direct computation:

$(2.19)^{3.5} \approx 10.48$

$a \cdot 10.48 = 21$

$a = \frac{21}{10.48} \approx 2.00$

Step 4: Finding $f(8)$

$f(8) = 2.00 \cdot (2.19)^8$

Computing:

$(2.19)^8 \approx 389.37$

$f(8) \approx 2.00 \times 389.37$

$f(8) \approx 778.74$

Final Answer:

$\mathbf{778.74}$

Would you like a detailed breakdown of any step?

Related Questions:

1. How do you derive the exponential function from two given points?
2. What is the significance of the base $b$ in an exponential function?
3. How do logarithms help in solving exponential equations?
4. How does changing $a$ or $b$ affect the shape of an exponential graph?
5. How can we approximate exponential functions from real-world data?

Tip:

When solving for exponential functions, always take the ratio of function values to eliminate $a$, making it easier to solve for $b$.