Problem 116. Chad deposit $ 2800 in-a CD for tax deduction reasons, and to hopefully be able to buy a car when his eventually gives out. The CD has an annual interest rate of 4.5% that is compounded continuously. We It will take Z036 years saw in Definition 98, that this will give us an exponential growth function of the form P(t) = 2500e0.045«t It will take Z036 years %3D 1. How many years will have elapsed before the CD is worth $7000? for the CD to be worth7000 A-P.e(r+) 0,3979co.8461. log @) 7000~メ0 0.645 92.5=e0.04st log (2.5) lege o.04s:) 8.8431 Tos G) 20.36-t 2. Find an equation for the inverse function that gives the time t as a function of the amount in the account.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem 116.** Chad deposits $2800 in a CD for tax deduction reasons, and to hopefully be able to buy a car when he eventually gives out. The CD has an annual interest rate of 4.5% that is compounded continuously. We saw in Definition 98, that this will give us an exponential growth function of the form \( P(t) = 2500e^{0.045t} \).

1. **How many years will have elapsed before the CD is worth $7000?**

    - Using the formula \( A = Pe^{(rt)} \), we solve for \( t \):
      \[
      7000 = 2800e^{0.045t}
      \]
      \[
      \frac{7000}{2800} = e^{0.045t}
      \]
      \[
      2.5 = e^{0.045t}
      \]
      Taking the natural logarithm:
      \[
      \log(2.5) = 0.045t \log(e)
      \]
      \[
      t = \frac{\log(2.5)}{0.045}
      \]
      Solution: \( t \approx 20.36 \) years

    - Handwritten calculation confirms: It will take approximately 20.36 years for the CD to be worth $7000.

2. **Find an equation for the inverse function that gives the time \( t \) as a function of the amount in the account.**

    - Solve for \( t \) in terms of \( A \):
      \[
      A = Pe^{(rt)}
      \]
      \[
      \frac{A}{P} = e^{rt}
      \]
      \[
      \log\left(\frac{A}{P}\right) = rt
      \]
      \[
      t = \frac{\log\left(\frac{A}{P}\right)}{r}
      \]

**Note:** There are several scribbled calculations on the paper indicating steps to solve for \( t \). Some handwritten notes also attempt to calculate with constants like 0.3979 and 0.645.
Transcribed Image Text:**Problem 116.** Chad deposits $2800 in a CD for tax deduction reasons, and to hopefully be able to buy a car when he eventually gives out. The CD has an annual interest rate of 4.5% that is compounded continuously. We saw in Definition 98, that this will give us an exponential growth function of the form \( P(t) = 2500e^{0.045t} \). 1. **How many years will have elapsed before the CD is worth $7000?** - Using the formula \( A = Pe^{(rt)} \), we solve for \( t \): \[ 7000 = 2800e^{0.045t} \] \[ \frac{7000}{2800} = e^{0.045t} \] \[ 2.5 = e^{0.045t} \] Taking the natural logarithm: \[ \log(2.5) = 0.045t \log(e) \] \[ t = \frac{\log(2.5)}{0.045} \] Solution: \( t \approx 20.36 \) years - Handwritten calculation confirms: It will take approximately 20.36 years for the CD to be worth $7000. 2. **Find an equation for the inverse function that gives the time \( t \) as a function of the amount in the account.** - Solve for \( t \) in terms of \( A \): \[ A = Pe^{(rt)} \] \[ \frac{A}{P} = e^{rt} \] \[ \log\left(\frac{A}{P}\right) = rt \] \[ t = \frac{\log\left(\frac{A}{P}\right)}{r} \] **Note:** There are several scribbled calculations on the paper indicating steps to solve for \( t \). Some handwritten notes also attempt to calculate with constants like 0.3979 and 0.645.
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