A company buys a machine for $20 million. Four years later, the machine is worth $12 million. Assume the decline in value follows the exponential model.

Question

A) Find the exponential model

 

B) Use the model to determine the machine's value in another 2 years.

Answer 1

A company purchases a machine for $20 million, and after four years, its value decreases to $12 million. Assuming the machine’s value declines exponentially, the model can be expressed as V(t)=V0ektV(t) = V_0 e^{kt}V(t)=V0​ekt, where V0V_0V0​ is the initial value and k is the rate of depreciation. Substituting the known values gives 12=20e4k12 = 20e^{4k}12=20e4k, which simplifies to 0.6=e4k0.6 = e^{4k}0.6=e4k. Taking the natural logarithm of both sides yields k=ln⁡(0.6)4=−0.1277k = \frac{\ln(0.6)}{4} = -0.1277k=4ln(0.6)​=−0.1277. Therefore, the exponential depreciation model is V(t)=20e−0.1277tV(t) = 20e^{-0.1277t}V(t)=20e−0.1277t. To find the value of the machine after another two years (six years in total), substitute t=6t = 6t=6 into the model: V(6)=20e−0.1277(6)=20e−0.7662≈9.3V(6) = 20e^{-0.1277(6)} = 20e^{-0.7662} \approx 9.3V(6)=20e−0.1277(6)=20e−0.7662≈9.3. Thus, after six years, the machine’s value is approximately $9.3 million.