Investiment equation

nevaco

New member
Joined
Dec 22, 2015
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4
Hey guys, year is running out and I'm preparing my personal finances and doing some budgeting, but then I came across this problem:

how long do I have to wait until I reach a certain goal (amount), given I start with an initial value invested at a certain interest rate and that I add some money regularly to that investiment?

To make it easier, I assume that both interest and additional money is added only once per year.

I came up with the following equation:

g = ( i + ay ) ( 1 + r)^y

where
g = goal
i = initial amount
a = yearly additions
r = interest rate
y = time (years)

My first question is:

Is then my equation correct? (I know it is considering the additional money at the beginning of the year, but that's fine for me, as I just want a rough estimative)

My second question is:

How to solve this equation on y? I've already played a lot with it, and I can't remember how to do it. My problem is that 'y' appears not only on the exponent but also multiplied by 'a'. So when I try to solve it, i end up with y inside a ln() and I don't know then how to proceed.

Here are my steps so far:

(1+r)^y = g / (i + ay)

ln (1+r)^y = ln g/(i + ay)

y ln (1+r) = ln g/(i + ay)

y = (ln g/(i + ay)) / ln (1+r) .... I can remove y from the 'ln' on the second part of the equation?


Thank you very much for any help in advance!
 
Thanks. I paste my result for future referral

so letting k = (1 + i)^n


f = ak + d[k - 1] / i

f = ak + (dk - d) / i

fi - aki = dk - d

k(d + ak) = fi + d

k = (fi + d) / (d + ak)

replacing k

(1+i)^n = (fi + d) / (d + ak)

n Log (1+i) = Log [
(fi + d) / (d + ak)]

n =
Log [(fi + d) / (d + ak)] / Log (1+i)
 
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