To derive an exponential equation for the given table, we'll start by assuming a model of the form:
\[ y = ab^x \]
where:
- \( y \) is the dependent variable,
- \( x \) is the independent variable,
- \( a \) is a constant (the initial value when \( x = 0 \)),
- \( b \) is the base of the exponential function.
From your data:
- When \( x = 1 \), \( y = 24 \)
- When \( x = 2 \), \( y = 12 \)
- When \( x = 3 \), \( y = 6 \)
- When \( x = 4 \), \( y = 3 \)
We can first determine the ratio between the \( y \) values as \( x \) increases:
\[ \frac{y(2)}{y(1)} = \frac{12}{24} = \frac{1}{2} \]
\[ \frac{y(3)}{y(2)} = \frac{6}{12} = \frac{1}{2} \]
\[ \frac{y(4)}{y(3)} = \frac{3}{6} = \frac{1}{2} \]
From this consistent ratio, we can establish that \( b = \frac{1}{2} \).
Now, to find \( a \):
- Use one of the data points when \( x = 1 \) and \( y = 24 \):
\[ y = ab^x \implies 24 = a\left(\frac{1}{2}\right)^1 \]
This simplifies to:
\[ 24 = \frac{a}{2} \]
So,
\[ a = 24 \cdot 2 = 48. \]
Now we have both constants \( a \) and \( b \). Therefore, the exponential equation that models your data is:
\[ y = 48\left(\frac{1}{2}\right)^x. \]
You can write this on math paper as:
\[ \boxed{y = 48 \left(\frac{1}{2}\right)^x} \]