Question

The graph of an exponential function passes through the points 1,
1
16
and 3,
1
64
. Write an equation for the function in the form y=a(b)x.
y=

Answer 1

To find the equation of the exponential function in the form \( y = a(b)^x \) that passes through the points \( (1, \frac{1}{16}) \) and \( (3, \frac{1}{64}) \), we will use these points to set up a system of equations.

From the first point \( (1, \frac{1}{16}) \):
\[
\frac{1}{16} = a(b)^1 \quad \Rightarrow \quad ab = \frac{1}{16} \quad \text{(1)}
\]

From the second point \( (3, \frac{1}{64}) \):
\[
\frac{1}{64} = a(b)^3 \quad \Rightarrow \quad ab^3 = \frac{1}{64} \quad \text{(2)}
\]

Now we have the system of equations:
1. \( ab = \frac{1}{16} \)
2. \( ab^3 = \frac{1}{64} \)

We can substitute equation (1) into equation (2). From equation (1), we have:
\[
a = \frac{1}{16b}
\]

Substituting \( a \) into equation (2):
\[
\left(\frac{1}{16b}\right)b^3 = \frac{1}{64}
\]
\[
\frac{b^2}{16} = \frac{1}{64}
\]

Now, multiply both sides by 16:
\[
b^2 = \frac{16}{64} = \frac{1}{4}
\]

Taking the square root of both sides gives us:
\[
b = \frac{1}{2} \quad \text{(since b must be positive for growth)}
\]

Now, substitute \( b = \frac{1}{2} \) back into equation (1) to find \( a \):
\[
a \left(\frac{1}{2}\right) = \frac{1}{16}
\]
\[
a = \frac{1}{16} \cdot 2 = \frac{1}{8}
\]

Now that we have determined \( a \) and \( b \), the equation of the exponential function is:
\[
y = \frac{1}{8} \left(\frac{1}{2}\right)^x
\]

Alternatively, you can write it in the form:
\[
y = \frac{1}{8} \cdot 2^{-x}
\]

Thus, the final equation is:
\[
\boxed{y = \frac{1}{8} \left(\frac{1}{2}\right)^x}
\]