Question

A man buys a motorcycle. After t months t value $V is given by V = 10000e^(-pt), where p is a constant. The value of the motorcycle after 12 months expected to be $4000. Calculate:
1) the expected value of the motorcycle after 18 months
2) the age of the motorcycle, to the nearest month, when its expected value will be $1000.

WHAT I'VE DONE SO FAR:
i) $10,000
ii) Do I find the value of P using the first given (i.e. price at 12 months) and just substitute t with 18 afterwards?
iii) Do I find the value of P using the first given and just substitute V with $1000?

Answer 1

ii) \[V = 10 000e ^{-pt}\]
\[4000 = 10000e^{-12p}\]

How do I apply natural log to this equation?

Answer 2

@Aurora-Borealis, Thank you very much! :) I'm not actually sure how to apply this... I haven't seen this kind of method of answering before.

Answer 3

Sorry my bad, it's wrong. :\

Answer 4

V = 10000e^(-pt)


4000 = 10000e^(-p*12)


4000/10000 = e^(-12p)


0.4 = e^(-12p)


ln(0.4) = -12p


ln(0.4)/(-12) = p


0.07635756 = p


p = 0.07635756


So the equation is V = 10000e^(-0.07635756t)


-------------------------------------


a)


Plug in t = 18 and evaluate to find its value after 18 months


V = 10000e^(-0.07635756t)


V = 10000e^(-0.07635756*18)


V = 10000e^(-1.37443608)


V = 10000(0.252982217)


V = 2529.82217


So after 18 months, the value is $2,529.82

Answer 5

b) Plug in V = 1000 and solve for t


V = 10000e^(-0.07635756t)


1000 = 10000e^(-0.07635756t)


1000/10000 = e^(-0.07635756t)


0.1 = e^(-0.07635756t)


ln(0.1) = -0.07635756t


ln(0.1)/(-0.07635756) = t


30.1552995 = t


t = 30.1552995


So it will take roughly 30.1552995 months for the value to become $1,000

Answer 6

Thank you very, very much! :D

Answer 7

you're welcome