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question 1
which of the following exponential functions passes through the points (0, 18) and (30, 64/27)?
$f(x)=12(\frac{1}{2})^{x/3}$
$f(x)=12(2)^{x/3}$
$f(x)=18(\frac{5}{4})^{x/2}$
$f(x)=18(3)^{x/5}$
$f(x)=12(\frac{1}{3})^{x/2}$
$f(x)=18(\frac{2}{3})^{x/6}$

Explanation:

Step1: Check the point (0, 18)

For an exponential - function of the form \(y = a\cdot b^{x/c}\), when \(x = 0\), \(y=a\cdot b^{0}=a\). So, for the function to pass through the point \((0,18)\), the value of \(a\) must be 18. This eliminates the functions \(f(x)=12(\frac{1}{2})^{x/3}\), \(f(x)=12(2)^{x/3}\), and \(f(x)=12(\frac{1}{3})^{x/2}\) since their \(a\) - values are 12.

Step2: Check the point (30, \(\frac{64}{27}\)) for the remaining functions

We have three remaining functions: \(f(x)=18(\frac{5}{4})^{x/2}\), \(f(x)=18(3)^{x/5}\), and \(f(x)=18(\frac{2}{3})^{x/6}\).
For \(f(x)=18(\frac{5}{4})^{x/2}\), when \(x = 30\), \(f(30)=18\cdot(\frac{5}{4})^{30/2}=18\cdot(\frac{5}{4})^{15}
eq\frac{64}{27}\).
For \(f(x)=18(3)^{x/5}\), when \(x = 30\), \(f(30)=18\cdot3^{30/5}=18\cdot3^{6}
eq\frac{64}{27}\).
For \(f(x)=18(\frac{2}{3})^{x/6}\), when \(x = 30\), \(f(30)=18\cdot(\frac{2}{3})^{30/6}=18\cdot(\frac{2}{3})^{5}\).
We know that \((\frac{2}{3})^{5}=\frac{32}{243}\), and \(18\cdot\frac{32}{243}=\frac{18\times32}{243}=\frac{576}{243}=\frac{64}{27}\).

Answer:

\(f(x)=18(\frac{2}{3})^{x/6}\)