Question 1145264
How do I take the monthly payment formula for compound interest and solve for t?
The formula I am referring to is:

A=P(1+r/n)^(nt) + M((1+r/n)^(nt) -1)(n/r)

Where A= Future Value of the investment
P = Principle or Present Value of the investment
r = interest rate
n = number of times per year interest is compounded
t = the number of years it will take for the investment to mature
M = the monthly payment or regular contribution to the investment account

Thank you!

MM
<pre>In order to solve for t, we need to get that variable by ITSELF
{{{matrix(1,3, A, "=", P(1 + r/n)^(nt) + M((1 + r/n)^(nt)-1) * (n/r))}}}
{{{matrix(1,3, A, "=", P(1 + r/n)^(nt) + (Mn/r)((1 + r/n)^(nt)-1))}}} ------- Multiplying last 2 expressions
{{{matrix(1,3, A, "=", P(1 + r/n)^(nt) + (Mn/r)(1 + r/n)^(nt) - Mn/r)}}} --------- Distributing last expression
{{{matrix(1,3, A + Mn/r, "=", P(1 + r/n)^(nt) + (Mn/r)(1 + r/n)^(nt))}}} --------- Adding {{{Mn/r}}} to both sides
{{{matrix(1,3, A + Mn/r, "=", (P + Mn/r)(1 + r/n)^(nt))}}} ------------------ Factoring out right side
{{{matrix(1,3, (A + Mn/r)/(P + Mn/r), "=", (P + Mn/r)(1 + r/n)^(nt)/(P + Mn/r))}}} ---------------- Dividing both sides by {{{P + (Mn/r)}}}
{{{matrix(1,3, ((Ar + Mn)/r)/((Pr + Mn)/r), "=", cross((P + Mn/r))(1 + r/n)^(nt)/cross((P + Mn/r)))}}}
{{{matrix(1,3, ((Ar + Mn)/r) * (r/(Pr + Mn)), "=", (1 + r/n)^(nt))}}} ---------- Applying <b><u>KCF</b></u> to left side
{{{matrix(1,3, (Ar + Mn)/(Pr + Mn), "=", (1 + r/n)^(nt))}}} ----------------------- Cancelling and simplifying left-side fractions
{{{matrix(1,3, nt, "=", log((1 + r/n),((Ar + Mn)/(Pr + Mn))))}}} ------------------ Converting to <b><u>LOGS</b></u>
{{{highlight(highlight_green(highlight(matrix(1,3, t, "=", (log((1 + r/n),((Ar + Mn)/(Pr + Mn))))/n))))}}}
That's it! Quite GRUELING, I might add!!