
four times a year so every 1/4 year multiply by 1 + .075/4 = 1.01875
do that for n years or in other words 4n periods
so
3 = (1.01875)^4n
log 3 = 4n log 1.01875
I suspect you can take it from there.

Lovely! Thank you so much Damon! <3

You must be familiar and know by heart the compound interest formula:
Amount = principal(1 + i)^n , where i is the periodic interest rate expressed
as a decimal, and n is the number of interest periods.
in your case i = .075/4 = .01875
and n = ??
amount = 3, principal = 1
then : 1(1.01875)^n = 3
you will need to use logs to solve this problem.
Let me know what you get.

The Accumulated amount needs to triple...
Do you know how to "undo" an exponent?
AKA do you know how to use logarithms?
If not... I will show you without them : )

MsPi, No, I do not know how to use logarithms. thanks :)

Reiny I don't even recall learning interest as I moved schools in the middle of the year when we started studying them.

So if you call your principal 1, then to triple it, it would need to grow to 3.
Thus the accumulated amount in the account would need to grow to 3.
Given
A=P(1+i)^n
3 = 1(1 + 0.075/4)^4n
3=(1.01875)^4n
since you can't recall logarithms (and may have missed that section of learning due to your school move).
start with n=1
3=(1.01875)^4(1)
now sub the right hand side into your calculator and see if it equals three
but the right hand side equals 1.077135 when n=1,
so continue in this manner until the right hand side is equal to 3,
let's try n=10
(1.01875)^4(10)
=2.1023
So we are not yet at 3, and have to continue to increase the number of times the money is looked at.
Let me know what you get : )
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