F-LE Two Points Determine an Exponential Function II

The value of the function decreases from 3 to \(\frac{16}{27}\) by multiplying 3 four times by \(b\) $$ \begin{align} 3 \times b \times b \times b \times b =& \frac{16}{27} \ 3b^4 =& \frac{16}{27} \ b^4 =& \frac{16}{81} \ b =& \pm \frac{2}{3} \end{align} $$

The base, \(b\), must be positive so \(b = \frac{2}{3}\)

Substituting the point \((-1, 3)\) yields

$$ \begin{align} 3=&a\frac{2}{3}^{-1} \text{ so} \ 3=&a\frac{3}{2} \text{ and} \ 2=&a \end{align} $$

So \(f(x) = 2\left(\frac{2}{3}\right)^x\)

Substituting the point \((-1, 3)\) yields \(3=ab^{-1}\), so \(3b=a\)

Substituting the point \((3, \frac{16}{27})\) yields $$\begin{align}\frac{16}{27} =& ab^3 \text{ so} \ \frac{16}{27} =& (3b)b^3 \text{ or} \ \frac{16}{27} =& 3b^4 \text{ and} \ \frac{16}{81} =& b^4 \ \pm \frac{2}{3} =& b\end{align}$$

The base, \(b\), must be positive so \(\frac{2}{3}=b\) and since \(3b=a\), \(3\left(\frac{2}{3}\right)=a\), or \(a=2\).

So \(f(x) = 2\left(\frac{2}{3}\right)^x\).