Answer :
We start with the exponential function
[tex]$$
f(x) = a b^x,
$$[/tex]
and we are given that
[tex]$$
f(4.5) = a\, b^{4.5} = 17 \quad \text{and} \quad f(5) = a\, b^5 = 94.
$$[/tex]
Step 1. Eliminate [tex]\(a\)[/tex] to find [tex]\(b\)[/tex]:
Divide the second equation by the first:
[tex]$$
\frac{a\, b^5}{a\, b^{4.5}} = \frac{94}{17}.
$$[/tex]
This simplifies to
[tex]$$
b^{5-4.5} = b^{0.5} = \frac{94}{17}.
$$[/tex]
Squaring both sides to solve for [tex]\(b\)[/tex],
[tex]$$
b = \left(\frac{94}{17}\right)^2.
$$[/tex]
Step 2. Solve for [tex]\(a\)[/tex]:
Use the equation for [tex]\(f(4.5)\)[/tex]:
[tex]$$
a\, b^{4.5} = 17.
$$[/tex]
Thus,
[tex]$$
a = \frac{17}{b^{4.5}}.
$$[/tex]
Step 3. Find [tex]\(f(7)\)[/tex]:
We substitute into the function:
[tex]$$
f(7) = a\, b^7.
$$[/tex]
Using our expression for [tex]\(a\)[/tex] we can write:
[tex]$$
f(7) = \frac{17}{b^{4.5}} \, b^7 = 17 \, b^{7-4.5} = 17 \, b^{2.5}.
$$[/tex]
Upon calculating the numerical value, we obtain
[tex]$$
f(7) \approx 87870.59810107638.
$$[/tex]
Rounding to the nearest hundredth, we have
[tex]$$
f(7) \approx 87870.60.
$$[/tex]
Thus, the value of [tex]\(f(7)\)[/tex] is
[tex]$$
\boxed{87870.60}.
$$[/tex]
[tex]$$
f(x) = a b^x,
$$[/tex]
and we are given that
[tex]$$
f(4.5) = a\, b^{4.5} = 17 \quad \text{and} \quad f(5) = a\, b^5 = 94.
$$[/tex]
Step 1. Eliminate [tex]\(a\)[/tex] to find [tex]\(b\)[/tex]:
Divide the second equation by the first:
[tex]$$
\frac{a\, b^5}{a\, b^{4.5}} = \frac{94}{17}.
$$[/tex]
This simplifies to
[tex]$$
b^{5-4.5} = b^{0.5} = \frac{94}{17}.
$$[/tex]
Squaring both sides to solve for [tex]\(b\)[/tex],
[tex]$$
b = \left(\frac{94}{17}\right)^2.
$$[/tex]
Step 2. Solve for [tex]\(a\)[/tex]:
Use the equation for [tex]\(f(4.5)\)[/tex]:
[tex]$$
a\, b^{4.5} = 17.
$$[/tex]
Thus,
[tex]$$
a = \frac{17}{b^{4.5}}.
$$[/tex]
Step 3. Find [tex]\(f(7)\)[/tex]:
We substitute into the function:
[tex]$$
f(7) = a\, b^7.
$$[/tex]
Using our expression for [tex]\(a\)[/tex] we can write:
[tex]$$
f(7) = \frac{17}{b^{4.5}} \, b^7 = 17 \, b^{7-4.5} = 17 \, b^{2.5}.
$$[/tex]
Upon calculating the numerical value, we obtain
[tex]$$
f(7) \approx 87870.59810107638.
$$[/tex]
Rounding to the nearest hundredth, we have
[tex]$$
f(7) \approx 87870.60.
$$[/tex]
Thus, the value of [tex]\(f(7)\)[/tex] is
[tex]$$
\boxed{87870.60}.
$$[/tex]
