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. A graph titled Light Intensity versus Distance has log (distance) on the x-axis, and log (Intensity) on the y-axis. Points curve down exponentially.

A 5-column table with 2 rows. Column 1 is labeled Predictor with entries constant, Log distance. Column 2 is labeled coefficient with entries 0.8561, negative 1.4966. Column 3 is labeled S E Coefficient with entries 0.1282, 0.0962. Column 4 is labeled T with entries 6.68, negative 15.56. Column 5 is labeled P with entries 0.001, 0.000. S = 0.0543, r square = 96.4 percent.

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.