The problem states that a laptop is initially valued at $2350. Each year, the laptop loses 1/5 of its value. We want to find the value of the laptop after 3 years.

AlgebraExponential DecayPercentageWord Problem
2025/3/26

1. Problem Description

The problem states that a laptop is initially valued at $
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0. Each year, the laptop loses 1/5 of its value. We want to find the value of the laptop after 3 years.

2. Solution Steps

Let V0V_0 be the initial value of the laptop. So, V0=2350V_0 = 2350.
Each year the laptop loses 1/5 of its value. This means that each year, the laptop retains 115=451 - \frac{1}{5} = \frac{4}{5} of its value.
After 1 year, the value of the laptop is V1=V0×45V_1 = V_0 \times \frac{4}{5}.
After 2 years, the value of the laptop is V2=V1×45=V0×(45)2V_2 = V_1 \times \frac{4}{5} = V_0 \times (\frac{4}{5})^2.
After 3 years, the value of the laptop is V3=V2×45=V0×(45)3V_3 = V_2 \times \frac{4}{5} = V_0 \times (\frac{4}{5})^3.
Plugging in the value of V0V_0:
V3=2350×(45)3=2350×4353=2350×64125=2350×64125=150400125=1203.2V_3 = 2350 \times (\frac{4}{5})^3 = 2350 \times \frac{4^3}{5^3} = 2350 \times \frac{64}{125} = \frac{2350 \times 64}{125} = \frac{150400}{125} = 1203.2.
The value of the laptop after 3 years is $1203.
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3. Final Answer

$1203.20