How can we isolate the $r$ in the following formula?
$$A = P\left(1 + \frac rn\right)^{nt} + \frac{PMT\left(\left(1 + \frac rn\right)^{nt} - 1\right)}{\frac rn}$$
I went through multiple tutorials but found it hard to solve. It will be really helpful if anyone can describe the steps for isolating the r in this formula. Thanks in advance.
On
There is no way to "isolate" $r$ but we can approximate it. The equation being $$A = P\left(1 + \frac rn\right)^{nt} + \frac{PMT\left(\left(1 + \frac rn\right)^{nt} - 1\right)}{\frac rn}$$ let $$a=\frac PA\qquad b=MT \qquad c=nt \qquad x=\frac r n$$ to make it $$a=\frac{x}{(b+x) (x+1)^c-b}$$ and we know that $x \ll 1$.
Developing the rhs as a Taylor series built at $x=0$, we have $$a=\alpha+\beta x+\gamma x^2+\delta x^3+O\left(x^4\right)$$ $$\alpha=\frac{1}{b c+1}\qquad \beta=-\frac{c (b(c-1)+2)}{2 (b c+1)^2}\qquad \gamma=\frac{c (c+1) (b (c-1) (b c+4)+6)}{12 (b c+1)^3}$$ $$\delta=-\frac{c (c+1) (b (c-1) (c (b (b c+4)+1)+6)+4 (c+2))}{24 (b c+1)^4} $$Now, using series reversion $$x=y-\frac{\gamma }{\beta } y^2+\frac{2 \gamma ^2-\beta \delta }{\beta ^2}y^3 +O\left(y^4\right) \qquad \text{where} \qquad y=\frac{a-\alpha }{\beta }$$
It is easy to get more terms in $y^m$ and have better and better approximations for $x$.
Let $x:=1+\dfrac rn$ and let us rename the coefficients. The equation reads
$$(1-px^m)(x-1)=q(x^m-1)$$ and is of the polynomial type if $nt$ is integer. It can have several roots and has no closed-form solution in general.