For context I am working trying to work out a formula based on the payout annuity formulas. With these forums I have been able to derive the expected term for a given payment:
Standard payout annuity for amount: $A = \frac{P(1-(1+\frac{R}{n})^{-nt})}{\frac{R}{n}}$
Solved for Time: $t = \frac{-ln(1-\frac{AR}{nP})}{nln{1+\frac{R}{n}}}$
Furthermore the Interest can be calculated as: $I = ntP-A$
I am looking to Solve the payment $P$ where in terms of a desired interest. Putting the two equations together and combining constants for simplicity I get:
$P = \frac{c}{n}\frac{1}{t}$ and $\frac{1}{t} = \frac{nln(b)}{-ln(1-aP)}$ where $a = \frac{AR}{n}$, $b = 1+\frac{R}{n}$, and $c = I-A$
I can simplify the equation into the following form:
$P = \frac{cln(b)}{-ln(1-\frac{a}{P})} \to 0 = cln(b)+Pln(1-\frac{a}{P})$ exponentiating
$e^{0} = e^{ln(cb)+Pln(1-\frac{a}{P})} = e^{ln(b^{c})}e^{ln{(1-\frac{a}{P})}^P}$
$(1-\frac{a}{P})^{P} = d$ where $d = \frac{1}{b^{c}}$
How can I evaluate the need P from this form?