two different cars each depreciate to 60% of their respective original values. the first car depreciates at…

Explanation:

Step1: Use depreciation formula

The depreciation formula is $A = P(1 - r)^t$, where $A$ is the final value, $P$ is the initial value, $r$ is the rate of depreciation, and $t$ is the time. Given $A = 0.6P$, for the first car with $r_1=0.1$, we have $0.6P=P(1 - 0.1)^t_1$. Canceling out $P$ on both sides, we get $0.6 = 0.9^{t_1}$. Taking the natural - logarithm of both sides: $\ln(0.6)=\ln(0.9^{t_1})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(0.6)=t_1\ln(0.9)$. So, $t_1=\frac{\ln(0.6)}{\ln(0.9)}\approx\frac{- 0.5108}{-0.1054}\approx4.85$ years.

Step2: Calculate time for second car

For the second car with $r_2 = 0.15$, we have $0.6P=P(1 - 0.15)^t_2$. Canceling out $P$ on both sides, we get $0.6 = 0.85^{t_2}$. Taking the natural - logarithm of both sides: $\ln(0.6)=\ln(0.85^{t_2})$. Using the property $\ln(a^b)=b\ln(a)$, we have $\ln(0.6)=t_2\ln(0.85)$. So, $t_2=\frac{\ln(0.6)}{\ln(0.85)}\approx\frac{-0.5108}{-0.1625}\approx3.14$ years.

Step3: Find the difference in ages

The difference $\Delta t=t_1 - t_2\approx4.85 - 3.14 = 1.71\approx1.7$ years.

Answer:

1.7 years