Answer
in the year 2024
Work Step by Step
The model we arrived at was $A=5000(1.00025)^{12t}$
(amount $t$ years after November 2010).
We now want to find $t$ for which
$\left[\begin{array}{ll}
5200 & =5000(1.00025)^{12t}\\
5200/5000 & =(1.00025)^{12t}\\
\log(52/50) & =12t\cdot\log 1.00025\\
t & =\dfrac{\log(52/50)}{12\log 1.00025}
\end{array}\right]$
$t\approx 13.075$
So, this will happen during the fourteenth year after 2010, in the year 2024.
