Question

as an object travels away from a light source, the intensity of the light on the object diminishes. to measure the influence of distance on light intensity, a student uses a light meter to record intensity, in lumens, from a source at various distances. the results, which compare distance in centimeters to the recorded light intensity, are shown in the scatterplot. to develop a linear model, the student next took the log of each distance and the log of each intensity and used computer software to find a least - square equation, shown in the computer output. using the computer output, the best estimate of the light intensity at 19 centimeters is: 0.0876, because 0.8561 - 1.4966(log 19)= - 1.058, and 10^{-1.058}=0.0876 lumens. 0.3472, because 0.8561 - 1.4966(log 19)= - 1.058, and e^{-1.058}=0.3472 lumens. 0.3964, because 0.8561(log 19)-1.4966 = - 0.4018, and 10^{-0.4018}=0.3964 lumens. 0.6691, because 0.8561(log 19)-1.4966 = - 0.4018, and e^{-0.4018}=0.6691 lumens.

Explanation:

Step1: Recall the linear model for log - transformed data

The linear model for the log - transformed data is of the form \(\log y=a + b\log x\). Here, \(x\) is the distance and \(y\) is the light intensity.

Step2: Calculate \(\log(19)\)

We know that \(\log(19)\approx1.2788\)

Step3: Substitute into the equation \(\log y=a + b\log x\)

Given \(a = 0.8561\) and \(b=-1.4966\), we substitute \(x = 19\) (so \(\log x=\log(19)\approx1.2788\)) into \(\log y=0.8561-1.4966\log x\)
\(\log y=0.8561-1.4966\times1.2788\)
\(\log y=0.8561 - 1.914\)
\(\log y=- 1.058\)

Step4: Convert from \(\log y\) to \(y\)

If \(\log y=-1.058\), then by the definition of the common logarithm (\(y = 10^{\log y}\)), we have \(y = 10^{-1.058}\)
Using the property of exponents \(a^{-n}=\frac{1}{a^{n}}\), \(10^{-1.058}=\frac{1}{10^{1.058}}\approx0.0876\)

Answer:

0.0876, because \(0.8561−1.4966(\log19)= - 1.058\), and \(10^{-1.058}=0.0876\) lumens.