Decay

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June 10th 2007, 11:45 AM
Danielisew

Decay

I am stuck on question 17 on this paper. Here goes...

Gwen bought a new car.
Each year, the value of her car depreciated by 9%

Calculate the number of years after which the value of her car was 47% of its value when new.

I just don't get this question at all.. Anyone help?

Thanks :)
June 10th 2007, 12:10 PM
topsquark

Quote:

Originally Posted by Danielisew

I am stuck on question 17 on this paper. Here goes...

Gwen bought a new car.
Each year, the value of her car depreciated by 9%

Calculate the number of years after which the value of her car was 47% of its value when new.

I just don't get this question at all.. Anyone help?

Thanks :)

Say the initial value of the car is $x_0$ . Then, for t in terms of years, the value x(t) of the car is
$x(t) = x_0(1 - 0.09)^t = x_0(0.91)^t$

So when the car has a value of 47% of its inital value, $x(t) = 0.47x_0$

Thus
$0.47x_0 = x_0(0.91)^t$

$0.47 = (0.91)^t$

Take the natural log (or log to any base in fact) of both sides:
$ln(0.47) = ln[(0.91)^t]$

$ln(0.47) = t \cdot ln(0.91)$

$t = \frac{ln(0.47)}{ln(0.91)} = 8.0057$

or about 8 years.

-Dan
June 10th 2007, 12:29 PM
Soroban

Hello, Danielisew!

Quote:

Gwen bought a new car.
Each year, the value of her car depreciated by 9%

Calculate the number of years after which the value of her car
is 47% of its original value.

Let's take baby-steps . . .

The original value is $V$ .

During the first year, the car loses 9% of its value.
. . Hence, its value is: . $v_1 \:=\:0.91V$

During the second year, the car loses 9% of that value.
. . Hence, its value is: . $v_2\:=\:(0.91)(0.91V) \:=\:(0.91)^2V$

During the third year, the car loses 9% of that value.
. . Hence, its value is: . $v_3\:=\:(0.91)(0.91^2V) \:=\:(0.91)^3V$

In general, after $n$ years, the car's value is: . $v_n\:=\:(0.91)^nV$

When is this 47% of the original value?

. . $\frac{(0.91)^nV}{V} \:=\:47\%\quad\Rightarrow\quad (0.91)^n \:=\:0.47$

Take logs: . $\log(0.91^n) \:=\:\log(0.47)\quad\Rightarrow\quad n\!\cdot\!\log(0.91) \:=\:\log(0.47)$

Then: . $n \:=\:\frac{\log(0.47)}{\log(0.91)} \:=\:8.005695522\;\cdots \text{ about 8 years}$