Use the exponential regression equation that best fits the data (10, 4), (12, 20), (13, 35), and (16, 300) to estimate $y$ when $x=15$.
We are given four data points: \((10, 4)\), \((12, 20)\), \((13, 35)\), and \((16, 300)\). Our goal is to fit an exponential model of the form \(y = ab^x\) to this data and use it to predict the value of \(y\) when \(x = 15\). To solve this, we can linearize the exponential model by taking the natural logarithm of both sides:
\[\ln(y) = \ln(ab^x) = \ln(a) + x \ln(b)\]
Letting \(Y = \ln(y)\), \(A = \ln(a)\), and \(B = \ln(b)\), we obtain a linear equation \(Y = A + Bx\). We can then perform a linear regression to find the constants \(A\) and \(B\) that best fit the data.
We transform our data \((x, y)\) into \((x, Y)\) where \(Y = \ln(y)\):
Using the method of least squares to fit the linear model \(Y = A + Bx\) to these four points, we determine the coefficients:
\(A \approx -5.679\)
\(B \approx 0.713\)
Now that we have the linear regression parameters \(A\) and \(B\), we can estimate \(y\) for \(x=15\) using the model:
\[Y = A + Bx\]
Substituting \(x=15\), \(A \approx -5.679\), and \(B \approx 0.713\):
\[Y = -5.679 + (0.713 \times 15) = -5.679 + 10.695 = 5.016\]
Since \(Y = \ln(y)\), we have:
\[y = e^{Y} = e^{5.016} \approx 150.56\]
Among the given options, \(146.7\) is the closest estimate.
Exponential models are frequently used in fields like finance and biology to track rapid growth. For instance, if you are tracking the population of a rapidly reproducing bacteria colony, you might collect data points at specific intervals and use exponential regression to forecast when the population will reach a certain size. This helps in understanding growth trends, making informed predictions about future resource needs, or estimating critical thresholds.