Question
upstudy study bank question image url

A computer purchased for \( \$ 1,200 \) loses \( 12 \% \) of its value every year. The computer's value can be modeled by the function \( v(t)=a \cdot b^{t} \), where \( v \) is the dollar value and \( t \) the number of years since purchase. (A) In the exponential model \( a=\square \) and \( b= \) (B) In how many years will the computer be worth half its original value? Round answer to 1 decimal place. The answer is Question 7

Ask by Sullivan Gardner. in the United States
Mar 13,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

**(A)** \( a = 1200 \) \( b = 0.88 \) **(B)** The computer will be worth half its original value in **5.4 years**.

Solution

**(A)** The initial value is \( a = 1200 \), and since the computer loses \( 12\% \) of its value each year, it retains \( 100\% - 12\% = 88\% \) of its value annually. Thus, the multiplicative factor is \[ b = 0.88. \] **(B)** We want to find the number of years \( t \) when the computer is worth half its original value, i.e., when \[ v(t) = 1200 \cdot 0.88^t = \frac{1200}{2} = 600. \] Dividing both sides of the equation by \( 1200 \) gives: \[ 0.88^t = \frac{1}{2}. \] Taking the natural logarithm of both sides: \[ \ln(0.88^t) = \ln\left(\frac{1}{2}\right). \] Using the logarithmic property \(\ln(x^y)= y\ln(x)\), we have: \[ t \ln(0.88) = \ln\left(\frac{1}{2}\right). \] Solving for \( t \): \[ t = \frac{\ln\left(\frac{1}{2}\right)}{\ln(0.88)}. \] Calculating the logarithms: \[ \ln\left(\frac{1}{2}\right) \approx -0.6931, \quad \ln(0.88) \approx -0.1278. \] Thus: \[ t \approx \frac{-0.6931}{-0.1278} \approx 5.42. \] Rounded to one decimal place, the computer will be worth half its original value in **5.4 years**.

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

The Deep Dive

To set up the exponential model, we have the initial value \( a = 1200 \) since that’s the price paid for the computer. The value decreases by \( 12\% \) each year, meaning it retains \( 88\% \) of its value, which can be represented as the decimal \( b = 0.88 \). Now, don't be discouraged; let's tackle when the computer will be worth half its original value. Half of the original \( \$1200 \) is \( \$600 \). We can set the equation \( v(t) = 1200 \cdot 0.88^t \), and we want to find \( t \) such that \( 1200 \cdot 0.88^t = 600 \). Solving for \( t \) gives: \[ 0.88^t = \frac{600}{1200} = 0.5 \] Taking the logarithm of both sides: \[ t \cdot \log(0.88) = \log(0.5) \] Calculating \( t \): \[ t = \frac{\log(0.5)}{\log(0.88)} \approx \frac{-0.3010}{-0.0530} \approx 5.7 \] So, it will take approximately \( 5.7 \) years for the computer to lose half its value! Now, for some fun tidbits! Did you know that the depreciation of electronics can often be much steeper than that of other types of assets? In fact, a new smartphone can lose up to \( 30\% \) of its value as soon as you unbox it! So if you're in the market for tech, timing your purchase wisely can save you a bundle! Additionally, there's a fascinating world behind how companies decide product prices and their depreciation rates! Many factors play into it, including market demand, material quality, and technological advances. If you're curious about the economics of technology, books on behavioral economics provide insight into consumer choices and price elasticity which reflect how we value gadgets over time!

Related Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy