Answer :
We are given an exponential function of the form
[tex]$$
f(x) = a \, b^x,
$$[/tex]
with the two pieces of information:
[tex]$$
f(4.5) = a \, b^{4.5} = 20 \quad \text{and} \quad f(11) = a \, b^{11} = 76.
$$[/tex]
Step 1. Solve for the Base [tex]$b$[/tex]:
Divide the equation for [tex]$f(11)$[/tex] by the equation for [tex]$f(4.5)$[/tex] to eliminate [tex]$a$[/tex]:
[tex]$$
\frac{a \, b^{11}}{a \, b^{4.5}} = \frac{76}{20}.
$$[/tex]
This simplifies to:
[tex]$$
b^{11 - 4.5} = \frac{76}{20}.
$$[/tex]
Since [tex]$11 - 4.5 = 6.5$[/tex], we have:
[tex]$$
b^{6.5} = \frac{76}{20} = \frac{19}{5}.
$$[/tex]
Taking the [tex]$6.5$[/tex]-th root gives:
[tex]$$
b = \left(\frac{19}{5}\right)^{\frac{1}{6.5}}.
$$[/tex]
Step 2. Solve for the Coefficient [tex]$a$[/tex]:
Substitute the value of [tex]$b$[/tex] into the equation for [tex]$f(4.5)$[/tex]:
[tex]$$
a \, b^{4.5} = 20 \quad \Longrightarrow \quad a = \frac{20}{b^{4.5}}.
$$[/tex]
Step 3. Compute [tex]$f(16.5)$[/tex]:
With [tex]$a$[/tex] and [tex]$b$[/tex] known, find [tex]$f(16.5)$[/tex]:
[tex]$$
f(16.5) = a \, b^{16.5}.
$$[/tex]
Substitute in the expression for [tex]$a$[/tex]:
[tex]$$
f(16.5) = \frac{20}{b^{4.5}} \cdot b^{16.5} = 20 \, b^{16.5 - 4.5} = 20 \, b^{12}.
$$[/tex]
After performing the computations using the values above, the base [tex]$b$[/tex] is approximately
[tex]$$
b \approx 1.227997482323005,
$$[/tex]
and the coefficient [tex]$a$[/tex] is approximately
[tex]$$
a \approx 7.93672535048231.
$$[/tex]
Thus,
[tex]$$
f(16.5) = a \, b^{16.5} \approx 235.17963526576344.
$$[/tex]
Rounding to the nearest hundredth results in:
[tex]$$
f(16.5) \approx 235.18.
$$[/tex]
Final Answer:
[tex]$$
\boxed{235.18}
$$[/tex]
[tex]$$
f(x) = a \, b^x,
$$[/tex]
with the two pieces of information:
[tex]$$
f(4.5) = a \, b^{4.5} = 20 \quad \text{and} \quad f(11) = a \, b^{11} = 76.
$$[/tex]
Step 1. Solve for the Base [tex]$b$[/tex]:
Divide the equation for [tex]$f(11)$[/tex] by the equation for [tex]$f(4.5)$[/tex] to eliminate [tex]$a$[/tex]:
[tex]$$
\frac{a \, b^{11}}{a \, b^{4.5}} = \frac{76}{20}.
$$[/tex]
This simplifies to:
[tex]$$
b^{11 - 4.5} = \frac{76}{20}.
$$[/tex]
Since [tex]$11 - 4.5 = 6.5$[/tex], we have:
[tex]$$
b^{6.5} = \frac{76}{20} = \frac{19}{5}.
$$[/tex]
Taking the [tex]$6.5$[/tex]-th root gives:
[tex]$$
b = \left(\frac{19}{5}\right)^{\frac{1}{6.5}}.
$$[/tex]
Step 2. Solve for the Coefficient [tex]$a$[/tex]:
Substitute the value of [tex]$b$[/tex] into the equation for [tex]$f(4.5)$[/tex]:
[tex]$$
a \, b^{4.5} = 20 \quad \Longrightarrow \quad a = \frac{20}{b^{4.5}}.
$$[/tex]
Step 3. Compute [tex]$f(16.5)$[/tex]:
With [tex]$a$[/tex] and [tex]$b$[/tex] known, find [tex]$f(16.5)$[/tex]:
[tex]$$
f(16.5) = a \, b^{16.5}.
$$[/tex]
Substitute in the expression for [tex]$a$[/tex]:
[tex]$$
f(16.5) = \frac{20}{b^{4.5}} \cdot b^{16.5} = 20 \, b^{16.5 - 4.5} = 20 \, b^{12}.
$$[/tex]
After performing the computations using the values above, the base [tex]$b$[/tex] is approximately
[tex]$$
b \approx 1.227997482323005,
$$[/tex]
and the coefficient [tex]$a$[/tex] is approximately
[tex]$$
a \approx 7.93672535048231.
$$[/tex]
Thus,
[tex]$$
f(16.5) = a \, b^{16.5} \approx 235.17963526576344.
$$[/tex]
Rounding to the nearest hundredth results in:
[tex]$$
f(16.5) \approx 235.18.
$$[/tex]
Final Answer:
[tex]$$
\boxed{235.18}
$$[/tex]
