Question

suppose the delinquency rate (as a percent) for consumer loans for a bank can be approximated by the function $f(x) = 8.26e^{-0.082x}$, where $x = 9$ corresponds to the year 2009. what was the first full year in which the delinquency rate fell below 2.75%? the first full year in which the delinquency rate fell below 2.75% was \\(\square\\) (type a whole number)

Explanation:

Step1: Set up the inequality

We need to find \( x \) such that \( 8.25e^{-0.082x} < 2.75 \). First, divide both sides by \( 8.25 \):
\( e^{-0.082x} < \frac{2.75}{8.25} \)
Simplify \( \frac{2.75}{8.25}=\frac{1}{3}\approx0.3333 \)
So we have \( e^{-0.082x} < 0.3333 \)

Step2: Take the natural logarithm of both sides

Since the natural logarithm is a monotonically increasing function, the inequality direction remains the same:
\( \ln(e^{-0.082x}) < \ln(0.3333) \)
Using the property \( \ln(e^a)=a \), we get:
\( -0.082x < \ln(0.3333) \)

Step3: Solve for \( x \)

First, calculate \( \ln(0.3333)\approx - 1.0986 \)
Then, divide both sides by \( - 0.082 \). Remember that dividing by a negative number reverses the inequality sign:
\( x > \frac{-1.0986}{-0.082} \)
Calculate \( \frac{1.0986}{0.082}\approx13.4 \)
So \( x > 13.4 \)

Step4: Determine the year

We know that \( x = 9 \) corresponds to 2009. So the year corresponding to \( x \) is \( 2009+(x - 9) \). Since \( x>13.4 \), the smallest integer \( x \) is 14.
Then the year is \( 2009+(14 - 9)=2014 \)

Answer:

2014