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?

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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?

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Answer 1 · 2015-06-14T11:47:11

The car will be worth $100 after 48 years and 23 days.

Explanation:

To decrease a number $x$ $x$ by 9.9%, you must calculate

$x \cdot (1 - \frac{9.9}{100}) = x \cdot 0.901$ $x*(1-9.9/100) = x*0.901$

Be $x_{0}$ $x_0$ the car's initial value, $x_{1}$ $x_1$ its value after one year, $x_{2}$ $x_2$ its value after two years, etc.

$x_{1} = x_{0} \cdot 0.901$ $x_1 = x_0*0.901$
$x_{2} = x_{1} \cdot 0.901 = x_{0} \cdot 0.901 \cdot 0.901 = x_{0} \cdot {(0.901)}^{2}$ $x_2=x_1*0.901=x_0*0.901*0.901=x_0*(0.901)^2$
$x_{y} = x_{0} \cdot {(0.901)}^{y}$ $x_y=x_0*(0.901)^y$ with $y$ $y$ the number of years that passed.

Therefore, the car's value on year $y$ $y$ is

$15000 {(0.901)}^{y}$ $15000(0.901)^y$

You want to know when the value will drop to $100, so you must solve this equation:

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

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

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

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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?

1 Answer

Explanation:

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