Given the principal sum P, the interest rate r, and the repayment M.
What is the formula for the number of repayments n?
Using this formula for calculating monthly repayments $$ M = \frac{Pr(1+r)^n}{ ((1 + r) ^n - 1)} $$
I rearranged as
$$ \frac{M}{Pr} = \frac{(1+r)^n}{((1+r)^n - 1)} $$
$$ (1+r)^n \frac{M}{Pr} - \frac{M}{Pr}= (1+r)^n $$
$$ \frac{- M}{Pr}= (1+r)^n - (1+r)^n \frac{M}{Pr} $$
$$ \frac{- M}{Pr}= (1+r)^n (1 -\frac{M}{Pr}) $$
$$ \frac{- M}{Pr -M}= (1+r)^n $$
$$ n = log_{1+r}( \frac{- M}{Pr -M})$$
$$ n = \frac{log_{10}( \frac{- M}{Pr -M})}{log_{10}(1+r)}$$
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Using this formula for calculating monthly repayments $$ M = \frac{Pr(1+r)^n}{ ((1 + r) ^n - 1)} $$
I rearranged as
$$ \frac{M}{Pr} = \frac{(1+r)^n}{((1+r)^n - 1)} $$
$$ (1+r)^n \frac{M}{Pr} - \frac{M}{Pr}= (1+r)^n $$
$$ \frac{- M}{Pr}= (1+r)^n - (1+r)^n \frac{M}{Pr} $$
$$ \frac{- M}{Pr}= (1+r)^n (1 -\frac{M}{Pr}) $$
$$ \frac{- M}{Pr -M}= (1+r)^n $$
$$ n = log_{1+r}( \frac{- M}{Pr -M})$$
$$ n = \frac{log_{10}( \frac{- M}{Pr -M})}{log_{10}(1+r)}$$