Question

a computer purchased for $550 loses 16% of its value every year. the computers value can be modeled by the function v(t) = a·b^t, where v is the dollar value and t the number of years since purchase.
(a) give the function that models the decrease in value of the computer: v(t) =
(b) in how many years will the computer be worth half its original value? round answer to 1 decimal place. years

Explanation:

Part (A)

Step1: Identify initial value \(a\)

The initial value of the computer is the purchase price, so \(a = 550\).

Step2: Determine the growth/decay factor \(b\)

Since the computer loses 16% of its value each year, the remaining value each year is \(100\% - 16\% = 84\% = 0.84\). So \(b = 0.84\).

Step3: Write the function

Using the formula \(v(t)=a\cdot b^{t}\), substitute \(a = 550\) and \(b = 0.84\) to get \(v(t)=550\cdot(0.84)^{t}\).

Step1: Set up the equation

The original value is 550, so half of the original value is \(\frac{550}{2}=275\). We need to solve \(550\cdot(0.84)^{t}=275\) for \(t\).

Step2: Divide both sides by 550

\(\frac{550\cdot(0.84)^{t}}{550}=\frac{275}{550}\), which simplifies to \((0.84)^{t}=0.5\).

Step3: Take the natural logarithm of both sides

\(\ln((0.84)^{t})=\ln(0.5)\). Using the logarithm power rule \(\ln(a^{b})=b\ln(a)\), we get \(t\ln(0.84)=\ln(0.5)\).

Step4: Solve for \(t\)

\(t=\frac{\ln(0.5)}{\ln(0.84)}\). Calculate \(\ln(0.5)\approx - 0.6931\) and \(\ln(0.84)\approx - 0.1744\). Then \(t=\frac{-0.6931}{-0.1744}\approx4.0\) (rounded to 1 decimal place).

Answer:

\(v(t) = 550 \cdot (0.84)^t\)

Part (B)