# The value of a car decreases at an annual rate of 9.9%. It is currently worth $15000. When will the car be worth$100?

Jun 14, 2015

The car will be worth $100 after 48 years and 23 days. #### Explanation: To decrease a number $x$by 9.9%, you must calculate $x \cdot \left(1 - \frac{9.9}{100}\right) = x \cdot 0.901$Be ${x}_{0}$the car's initial value, ${x}_{1}$its value after one year, ${x}_{2}$its value after two years, etc. ${x}_{1} = {x}_{0} \cdot 0.901$${x}_{2} = {x}_{1} \cdot 0.901 = {x}_{0} \cdot 0.901 \cdot 0.901 = {x}_{0} \cdot {\left(0.901\right)}^{2}$${x}_{y} = {x}_{0} \cdot {\left(0.901\right)}^{y}$with $y$the number of years that passed. Therefore, the car's value on year $y$is $15000 {\left(0.901\right)}^{y}$You want to know when the value will drop to$100, so you must solve this equation:

$15000 {\left(0.901\right)}^{y} = 100$
${0.901}^{y} = \frac{1}{150}$

Turn the power into a factor with the $\log$ function:
color(gray)(log(1)=0;log(a/b)=log(a)-log(b);log(a^b)=blog(a))

$\log \left({0.901}^{y}\right) = \log \left(\frac{1}{150}\right)$
$y \log \left(0.901\right) = - \log \left(150\right)$
$y = - \log \frac{150}{\log} \left(0.901\right) \approx 48.064$ years $\approx 48$ years and $23$ days