The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.
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What is depreciation?</h3>
Depreciation is to decrease in the value of a product in a period of time. This can be given as,
![FV=P\left(1-\dfrac{r}{100}\right)^n](https://tex.z-dn.net/?f=FV%3DP%5Cleft%281-%5Cdfrac%7Br%7D%7B100%7D%5Cright%29%5En)
Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.
Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is
. Thus, by the above formula for the first car,
![0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}](https://tex.z-dn.net/?f=0.6x%3Dx%5Cleft%281-%5Cdfrac%7B10%7D%7B100%7D%5Cright%29%5E%7Bn_1%7D%5C%5C0.6%3D%281-0.1%29%5E%7Bn_1%7D%5C%5C0.6%3D%280.9%29%5E%7Bn_1%7D)
Take log both the sides as,
![\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85](https://tex.z-dn.net/?f=%5Clog%200.6%3D%5Clog%20%280.9%29%5E%7Bn_1%7D%5C%5C%5Clog%200.6%3D%7Bn_1%7D%5Clog%20%280.9%29%5C%5Cn_1%3D%5Cdfrac%7B%5Clog%200.6%7D%7B%5Clog%200.9%7D%5C%5Cn_1%5Capprox4.85)
Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.
Thus, the depreciation price of the car is 0.6y. Let the number of year is
. Thus, by the above formula for the second car,
![0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}](https://tex.z-dn.net/?f=0.6y%3Dy%5Cleft%281-%5Cdfrac%7B15%7D%7B100%7D%5Cright%29%5E%7Bn_2%7D%5C%5C0.6%3D%281-0.15%29%5E%7Bn_2%7D%5C%5C0.6%3D%280.85%29%5E%7Bn_2%7D)
Take log both the sides as,
![\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14](https://tex.z-dn.net/?f=%5Clog%200.6%3D%5Clog%20%280.85%29%5E%7Bn_2%7D%5C%5C%5Clog%200.6%3D%7Bn_2%7D%5Clog%20%280.85%29%5C%5Cn_2%3D%5Cdfrac%7B%5Clog%200.6%7D%7B%5Clog%200.85%7D%5C%5Cn_2%5Capprox3.14)
The difference in the ages of the two cars is,
![d=4.85-3.14\\d=1.71\rm years](https://tex.z-dn.net/?f=d%3D4.85-3.14%5C%5Cd%3D1.71%5Crm%20years)
Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.
Learn more about the depreciation here;
brainly.com/question/25297296