Question

A sculpture was purchased for $4,000. The value of the sculpture increases at a rate of 4% year. How would I figout how much the sculpture would be worth after 10 years?

Answer 1

4000(1+4/100)^10
Now solve

Answer 2

About $5,921?

Answer 3

A{0} + A{0}(.04)

A{0} (1.04)

A{n} = A{n-1}(1.04)
= (A{n-2}(1.04))(1.04)
= ((A{n-3}(1.04))(1.04))(1.04)
= (A{n-r} (1.04)^r

when A{n-r} = A{0} when r=n

A{n} = (A{n-n} (1.04)^n
= A{0} (1.04)^n

A{0} = 4000 so...

A{n} = 4000(1.04^n); when n=10 is the value you want right?

Answer 4

Yeap

Answer 5

Whoa. Um, yes. So my calculation was correct?

Answer 6

Yes

Answer 7

Is amistre64 just breaking it down for me?

Answer 8

http://www.google.com/search?aq=f&sourceid=chrome&ie=UTF-8&q=4000(1.04^10)

id say yes :)

Answer 9

Yes, the formula comes from the recurssion

Answer 10

Thank you so much guys! :)

Answer 11

amistre Let's give him a break with the derivation, shall we? :-)

Answer 12

but..but... i need the practice lol

Answer 13

In how many years will the sculpture be worth $10,000? Round to the nearest tenth of a year. :/

Answer 14

10000=4000(1+4/100)^T

Answer 15

Now solve in the same way I helped u in the previous question

Answer 16

4160?

Answer 17

I don't think so, check it again

Answer 18

4000?

Answer 19

The sculpture was initially purchased for 4000 ,so

Answer 20

:/ I'm stuck. After 50 years?

Answer 21

lets view it like this:

A = P (1+r)^t ; and we solve for t, divide out the P

A/P = (1+r)^t ; now we go to logs to get the "t"

ln(A/P) = [ln(1+r)^t]

ln(A/P) = t ln(1+r) ; divide out the ln(1+r)

ln(A/P)
------ = t
ln(1+r)

Answer 22

Wait, what does A, P, and r stand for again?

Answer 23

Hello?