Question

a laptop computer is purchased for $4300. each year, its value is 80% of its value the year before. after how many years will the laptop computer be worth $1000 or less? (use the calculator provided if necessary.) write the smallest possible whole number answer. years

Explanation:

Step1: Define decay formula

The value of the laptop follows exponential decay: $V(t) = 4300 \times (0.8)^t$, where $V(t)$ is value after $t$ years. We need $V(t) \leq 1000$.

Step2: Set up inequality

$$4300 \times (0.8)^t \leq 1000$$

Step3: Isolate exponential term

Divide both sides by 4300:
$$(0.8)^t \leq \frac{1000}{4300} \approx 0.2326$$

Step4: Apply logarithm to solve t

Take natural log of both sides, reverse inequality since $\ln(0.8) < 0$:
$$t \geq \frac{\ln(0.2326)}{\ln(0.8)}$$
Calculate values: $\ln(0.2326) \approx -1.458$, $\ln(0.8) \approx -0.2231$
$$t \geq \frac{-1.458}{-0.2231} \approx 6.535$$

Step5: Round up to whole number

Since $t$ must be a whole number, round up to the next integer.

Answer:

7 years