Solving for a Special $n$

$\begin{aligned} 20^{2015} & = 2^n \times 40^9 \times 10^{2000} \\ 2^{2015} \times 10^{2015} & = 2^{n+18} \times 10^{2009} \\ \implies 2^n & = 2^{1997} \times 10^6 \\ n \log_{10}2 & = 1997 \log_{10} 2 + 6 \\ \implies n & = 1997 + \frac 6{\log_{10} 2} \\ & \approx 1997 + \frac 6{0.3010} \\ & \approx 2017 \\ \implies n+1 & \approx \boxed{2018} \end{aligned}$

Note: The title has been changed. So this is invalid.

How i soled it without calculating is that the title says that the answer is the year. The year is 2018.