Question

Can someone explain to me how I would go about doing part 2? (problem posted below)

I've been at this for days, and it's driving me nuts because I feel like I'm so close to the answer. I think the function will look something like this: assuming that t= years, f(t)=100 + .20(t)
or f(t)= 100 x .20(t), but I'm just not sure, and I think what I did is wrong anyway.

Can someone explain to me how to go about doing this?

Answer 1

So we start with, \(\Large\rm 100\).
After a year, the money has grown by 20%.
I like to think of this as 120% of the original amount (1.2 as a decimal) \(\Large\rm 100\cdot 1.2\)

After 2 years it has grown by another 20%, but of this new amount.
So it has increased to 120% of it's previous amount again: \(\Large\rm 100\cdot (1.2)^2\)

Answer 2

Do you see the pattern I was going for? :o

Answer 3

What do they mean by (actual)?
That is strange.. :P

Answer 4

Yes, I do see the pattern! And I don't know, I think that just means an exact rate of 20%. Or something along those lines . . . So, would I do something, say, like this?: f(t)=100 ⋅ (1.2)^t?

Answer 5

Mmmm yah I think that's what we're looking for!

Answer 6

Nothing nothing :) we're good.
I was trying to compare our solution to your original guess.
But I got a little mixed up.

Answer 7

Ohhh, ok xD But thank you so much for helping me, I really appreciate it!

Answer 8

np c:

Answer 9

that would be interest only right ?
the actual amount we would get is by adding $100 , isn;t it ?

Answer 10

@zepdrix , just had a small doubt ^

Answer 11

I think if we had something like \(\Large\rm f(t)=100(0.2)^t\) then THAT would only represent interest.

That's calculating a bunch of multiples of 20%.
I think the 1.2 makes it ok, yes?

Answer 12

like after 1 year $120 will be in your account

Answer 13

ok, plugging in t=1
gives 120
yes, its ok! :)
thanks for clarifying :)