Question

The value of a certain car depreciates according to v(t)=18500e^-0.186t where t is the number of years after the car is purchased new.
a) what will the car be worth in 18 months?
b) when will the car be worth half of its original value?

Answer 1

well what is 18 months in terms of years?
1 year = 12 months
2 years = 24 months
? years = 18 months?

looks like 1.5 years right? so plug that in for "t" in part A) and you'll get your answer

for part B) you see that the original value was 18500...so you want to find when the value is half that....so half of 18500 is 9250

so

9250 = 18500e^-0.186t

divide both sides by 18500

.5 = e^-0.186t

take the natural log of both sides

ln(.5) = -0.186t

and finally solve for t

ln(.5)
------ = t
-0.186

this will tell you after how many years the car will equal half of its value

Answer 2

I got a wicked weird number on my calculator..

Answer 3

It doesn't look right

Answer 4

for which one...and what number did you get?

Answer 5

part 2 I got .0918105205+.17...

Answer 6

ln(.5)
------ = t
-0.186

is what you should be doing right?

so

ln(.5) = -0.6931471805
and divide that by -0.186

this will give you "t"

Answer 7

Where did you get the number -0.6931471805?

Answer 8

that is what the natural log of 1/2 or .5 is

Answer 9

Thank you! I understand it now

Answer 10

are you sure? I can explain it better if need be...

Answer 11

Noo I get it! Would you mind helping me with another problem?

Answer 12

sure :)

Answer 13

Thanks :) okay so its -4ln2x=-26

Answer 14

I thought you had to make that \[\ln2x ^{-4}=-26\] and then divide by 2 so you would get \[lnx ^{-4}\] but then I got stuck

Answer 15

oops =13

Answer 16

Go back to your original post of that problem.

Answer 17

What?

Answer 18

You posted that problem earlier, you know.

Answer 19

Yeah but you didn't make sense when you explained it.

Answer 20

http://openstudy.com/study#/updates/515793a1e4b0507ceba1f0ab

Answer 21

I posted the solution.

Answer 22

Yeah and I still don't understand it