If $$ f(x) $$ is an exponential function of the form of $$ y=a b^{x} $$ where $$ f(4)=16 $$ and $$ f(4.5)=54 $$, then find the value of $$ f(5) $$, to the nearest hundredth.

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Given that $$ f(x) $$ is an exponential function of the form $$ y = ab^x $$, where $$ f(4) = 16 $$ and $$ f(4.5) = 54 $$, we need to find the value of $$ f(5) $$ to the nearest hundredth.

To find the value of $$ f(5) $$, we first need to determine the values of $$ a $$ and $$ b $$ using the given information.

From the given information, we have:
1. $$ f(4) = 16 $$
2. $$ f(4.5) = 54 $$

We can use these two equations to solve for $$ a $$ and $$ b $$.

Let's start by using the first equation $$ f(4) = 16 $$ to find the value of $$ a $$:
$$ 16 = ab^4 $$

Now, let's use the second equation $$ f(4.5) = 54 $$ to find the value of $$ b $$:
$$ 54 = ab^{4.5} $$

We can solve these two equations simultaneously to find the values of $$ a $$ and $$ b $$.

Let's solve these equations to find the values of $$ a $$ and $$ b $$ and then use them to find the value of $$ f(5) $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}16=ab^{4}\54=ab^{4.5}\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}16=ab^{4}\54=ab^{\frac{9}{2}}\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}a=\frac{16}{b^{4}}\54=ab^{\frac{9}{2}}\end{array}\right.$$
- step3: Substitute the value of $$a:$$
  $$54=\frac{16}{b^{4}}\times b^{\frac{9}{2}}$$
- step4: Reduce the fraction:
  $$54=16b^{\frac{1}{2}}$$
- step5: Swap the sides:
  $$16b^{\frac{1}{2}}=54$$
- step6: Divide both sides:
  $$\frac{16b^{\frac{1}{2}}}{16}=\frac{54}{16}$$
- step7: Divide the numbers:
  $$b^{\frac{1}{2}}=\frac{27}{8}$$
- step8: Raise both sides to the $$2$$-th power$$:$$
  $$\left(b^{\frac{1}{2}}\right)^{2}=\left(\frac{27}{8}\right)^{2}$$
- step9: Evaluate the power:
  $$b=\frac{729}{64}$$
- step10: Substitute the value of $$b:$$
  $$a=\frac{16}{\left(\frac{729}{64}\right)^{4}}$$
- step11: Simplify:
  $$a=\frac{2^{28}}{729^{4}}$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}a=\frac{2^{28}}{729^{4}}\b=\frac{729}{64}\end{array}\right.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}a=\frac{2^{28}}{729^{4}}\b=\frac{729}{64}\end{array}\right.$$
- step14: Rewrite:
  $$\left(a,b\right) = \left(\frac{2^{28}}{729^{4}},\frac{729}{64}\right)$$
The solution to the system of equations is $$ a = \frac{2^{28}}{729^{4}} $$ and $$ b = \frac{729}{64} $$.

Now that we have the values of $$ a $$ and $$ b $$, we can find the value of $$ f(5) $$ using the exponential function $$ f(x) = ab^x $$.

Substitute the values of $$ a $$ and $$ b $$ into the function to find $$ f(5) $$:
$$ f(5) = \frac{2^{28}}{729^{4}} \times \left(\frac{729}{64}\right)^5 $$

Let's calculate the value of $$ f(5) $$ to the nearest hundredth.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{2^{28}}{729^{4}}\times \left(\frac{729}{64}\right)^{5}$$
- step1: Evaluate the power:
  $$\frac{2^{28}}{729^{4}}\times \frac{729^{5}}{64^{5}}$$
- step2: Rewrite the expression:
  $$\frac{2^{28}}{729^{4}}\times \frac{729^{5}}{2^{30}}$$
- step3: Reduce the numbers:
  $$1\times \frac{729}{2^{2}}$$
- step4: Multiply:
  $$\frac{729}{2^{2}}$$
- step5: Evaluate the power:
  $$\frac{729}{4}$$
The value of $$ f(5) $$ to the nearest hundredth is 182.25.
$$ f(5) = 182.25 $$

If \( f(x) \) is an exponential function of the form of \( y=a b^{x} \) where \( f(4)=16 \) and \( f(4.5)=54 \), then find the value of \( f(5) \), to the nearest hundredth.

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