Question

two function models \\(g(t)\\) and \\(n(t)\\) are being considered to represent the sales for a company manufacturing graphing calculators. both \\(g(t)\\) and \\(n(t)\\) represent the sales of the calculators, in thousands of units, after \\(t\\) years for \\(9 \le t \le 15\\). if \\(g(t) = 11 - 2.144\ln(x)\\) and \\(n(t) = -x + 15\\), what is the first time \\(t\\) that sales predicted by the logarithmic model will be 0.5 thousand units more than sales predicted by the linear model?

9.275
9.881
10.432
12.116

Explanation:

Set up the equation

We seek the first time \(t\) where logarithmic sales exceed linear sales by \(0.5\).
Using Exponential Equations concepts, we set up the relation:

$$G(t) = N(t) + 0.5$$

Substituting the given functions (treating \(x\) in the text as the variable \(t\)):

$$11 - 2.144\ln(t) = -t + 15 + 0.5$$

Simplify the equation

Combine the constant terms on the right side:

$$11 - 2.144\ln(t) = -t + 15.5$$

Rearrange the terms to set the equation to zero:

$$t - 2.144\ln(t) - 4.5 = 0$$

Evaluate the options

We test the given multiple-choice options in the equation \(f(t) = t - 2.144\ln(t) - 4.5\):

For \(t = 9.275\):

$$f(9.275) = 9.275 - 2.144\ln(9.275) - 4.5 \approx 9.275 - 4.775 - 4.5 = 0$$

For \(t = 9.881\):

$$f(9.881) = 9.881 - 2.144\ln(9.881) - 4.5 \approx 9.881 - 4.911 - 4.5 = 0.47$$

For \(t = 10.432\):

$$f(10.432) = 10.432 - 2.144\ln(10.432) - 4.5 \approx 10.432 - 5.027 - 4.5 = 0.905$$

For \(t = 12.116\):

$$f(12.116) = 12.116 - 2.144\ln(12.116) - 4.5 \approx 12.116 - 5.348 - 4.5 = 2.268$$

Identify the correct time

The value \(t \approx 9.275\) yields approximately \(0\), satisfying the condition.

Answer:

• (A) 9.275 (Correct answer)
• (B) 9.881
• (C) 10.432
• (D) 12.116