How do I make $r$ the subject of formula? Comment: How to go about this question has been a Herculean task for me. 
I would be glad if an explanation is provided alongside with the calculation as regards making $ r $ the subject of formula in the equation: $$ pv = \frac{1}{r} - \frac{1}{r(r + 1)^t} $$
How far I have gone solving the problem: At first, I did take the LCM of the denominators at the RHS of the equation. I solved the problem to the point where I got stuck at 
$$ (r + 1)^t (1 - pvr) = 1 $$
 A: As told in comments, you cannot make $r$ the subject and some numerical method is required for finding the zero of function (for simplicity I changed notations : $p$ stands for $pv$)
$$f(r)=(r + 1)^t (1 - p\,r) - 1$$
The simplest to use is Newton method which, starting from an initial guess $r_0$ will update it according to 
$$r_{n+1}=r_n-\frac{f(r_n)}{f'(r_n)}$$ 
However, since $r$ is small $(r \ll 1)$, we can make approximations using series expansions around $r=\frac 1p$. This would give
$$f(r)=-1-p \left(r-\frac{1}{p}\right) \left(\frac{1}{p}+1\right)^t-p t
   \left(r-\frac{1}{p}\right)^2
   \left(\frac{1}{p}+1\right)^{t-1}+O\left(\left(r-\frac{1}{p}\right)^3\right)\tag1$$ Ignoring the $\left(r-\frac{1}{p}\right)^2$ term and higher powers of $r$, this would give an estimate
$$r_0=\frac1p-\frac{\left(\frac{1}{p}+1\right)^{-t}}{p}$$ For illustation puposes, let us try using $r=0.005$ and $t=180$; this would give $p=118.504$; reusing these numbers would lead to $r_0=0.00657919$ which is much too high. However, Newton iterations will be
$$\left(
\begin{array}{cc}
 n & r_n \\
 0 & 0.006579187739 \\
 1 & 0.005481598716 \\
 2 & 0.005061970447 \\
 3 & 0.005001209158 \\
 4 & 0.005000000473 \\
 5 & 0.005000000000
\end{array}
\right)$$ 
