A boat costs $11,850 and decreases in value by 10% per year. How much will the boat be worth after 8 years?
Could someone walk through with me?

Question

A boat costs $11,850 and decreases in value by 10% per year. How much will the boat be worth after 8 years?
Could someone walk through with me?

mathmale · Answer

The trick here is to recognize that this is a case of EXPONENTIAL DECAY.
An appropriate model would be $$y = a*e ^{rt}$$  That r (which represents the rate of decay) is negative here (it'd be positive for exponential growth).

There's more than one way in which to attack this problem.  One way is to recognize that the initial value of y (that is, the value of the boat) is $11850, so our model becomes$$y=$11850e ^{rt}$$

We're told that the value of the boat decreases at the rate of 10% per year.   Since 10% => 0.10, t$$y=$11850e ^{-0.10t}$$his translates into

mathmale · Answer

How much is the boat worth after 8 years?  Merely substitute 8 (years) for t and evaluate the resulting expression.

anonymous · Answer

You still there?

mathmale · Answer

Yes.  Have you tried finishing this calculation?

anonymous · Answer

I was looking at what you were saying and was wondering, where is the t? next to the y = 11850?

anonymous · Answer

Sorry, I'm all over the place and I'm exhausted.

anonymous · Answer

So 10% of 11850 is.. 1,185? Then we keep subtracting 1,185 from 11,850, 8 times??

whpalmer4 · Answer

No, the number subtracted is different each time.

Boat starts out being worth 11850.  

After 1 year, it loses 10% of its value, so it is now worth $11850 - (11850*0.1)$.  
If we factor that, we can write it as $11850(1-0.1) = 11850*0.9$.  

After another year, another 10% of the current value is lost, so after two years, the value of the boat is $(11850*0.9)*0.9 = 11850*(0.9)^2$.

After a 3rd year, another 10% of the current value is lost, so the value of the boat is
$11850*(0.9)^3$ and so on.  After $n$ years, the value will be $11850*(0.9)^n$

Compound interest works the same way, except there you are multiplying repeatedly by a number larger than 1, not smaller.

whpalmer4 · Answer

A handy rule of thumb for rough estimating is that 10% growth leads to a doubling in just over 7 years ($\approx 7.3$) .  10% decay leads to a halving in just under 7 years ($\approx 6.6$).