To determine the multiplicative rate of change of the given exponential function \( y = a \cdot r^x \), we look at the ratios of consecutive \( y \) values for given \( x \) values. This is because, in an exponential function, each term is the previous term multiplied by a constant rate, \( r \).
Given:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & \frac{3}{2} \\
\hline
2 & \frac{9}{8} \\
\hline
3 & \frac{27}{32} \\
\hline
4 & \frac{81}{128} \\
\hline
\end{array}
\][/tex]
To find \( r \), we calculate the ratio of \( y \) values using the following steps:
1. Calculate \( r \) from \( x = 1 \) and \( x = 2 \):
[tex]\[
r = \frac{y(2)}{y(1)} = \frac{\frac{9}{8}}{\frac{3}{2}} = \frac{9}{8} \times \frac{2}{3} = \frac{9 \cdot 2}{8 \cdot 3} = \frac{18}{24} = \frac{3}{4}
\][/tex]
2. Calculate \( r \) from \( x = 2 \) and \( x = 3 \):
[tex]\[
r = \frac{y(3)}{y(2)} = \frac{\frac{27}{32}}{\frac{9}{8}} = \frac{27}{32} \times \frac{8}{9} = \frac{27 \cdot 8}{32 \cdot 9} = \frac{216}{288} = \frac{3}{4}
\][/tex]
3. Calculate \( r \) from \( x = 3 \) and \( x = 4 \):
[tex]\[
r = \frac{y(4)}{y(3)} = \frac{\frac{81}{128}}{\frac{27}{32}} = \frac{81}{128} \times \frac{32}{27} = \frac{81 \cdot 32}{128 \cdot 27} = \frac{2592}{3456} = \frac{3}{4}
\][/tex]
In each case, we find that the ratio \( r \) is constant and equals \( \frac{3}{4} \).
Thus, the multiplicative rate of change of the function is \( \frac{3}{4} \).
The correct answer is:
[tex]\(\boxed{\frac{3}{4}}\)[/tex]
