Given that $f(x) = ab^x$ is an exponential function, and we have two points: $f(0.5) = 26$ and $f(4.5) = 31$, we need to find the value of $f(2.5)$ to the nearest hundredth.

Step 1: Set Up the System of Equations

We know: $f(0.5) = ab^{0.5} = 26$ $f(4.5) = ab^{4.5} = 31$

Step 2: Divide the Two Equations

By dividing $f(4.5)$ by $f(0.5)$, we eliminate $a$ and solve for $b$: $\frac{ab^{4.5}}{ab^{0.5}} = \frac{31}{26}$ This simplifies to: $b^{4.5 - 0.5} = \frac{31}{26}$ $b^4 = \frac{31}{26}$ $b = \left(\frac{31}{26}\right)^{\frac{1}{4}}$

Step 3: Calculate the Value of $b$

$b \approx \left(\frac{31}{26}\right)^{\frac{1}{4}} \approx 1.045$

Step 4: Find the Value of $a$

Substitute $b$ back into the equation $ab^{0.5} = 26$: $a \times (1.045)^{0.5} = 26$ $a \times 1.022 = 26$ $a \approx \frac{26}{1.022} \approx 25.43$

Step 5: Find $f(2.5)$

Now, substitute $x = 2.5$ into the function $f(x) = ab^x$: $f(2.5) = 25.43 \times (1.045)^{2.5}$ $f(2.5) \approx 25.43 \times 1.114 \approx 28.33$

Thus, $f(2.5) \approx 28.33$ to the nearest hundredth.

Would you like further details or have any questions?

Here are five related questions:

1. How would the solution change if $f(0.5)$ were different?
2. What is the general approach for solving exponential equations?
3. How can you verify the accuracy of the values for $a$ and $b$?
4. What role does logarithmic transformation play in solving exponential equations?
5. How would you determine $f(3)$ using the same method?

Tip: Always check the units and decimal places when rounding to ensure accuracy in final results.