Prove that $2002^{2001}$ and $2002^{2001}+2^{2001}$ have the same number of digits. I have an arithmetic exercise as follows: Prove that $2002^{2001}$ and $2002^{2001}+2^{2001}$ have a same number of digits. It seems easy but I don't know how to do. I can do it if $2002$ is replaced by $2000$. Any hints are appreciated.
Thanks a lot!

$\ln (a + b) = \ln (a(1 + \dfrac{b}{a})) = \ln a + \ln (1 + \dfrac{b}{a})$.
The number of digits in the base-10 numeral for integer $n > 0$ is $\lceil \log_{10}n \rceil$.  $\log_{10} n = \dfrac{\ln n}{\ln 10}$
Let $a = 2002^{2001}$ and $b = 2^{2001}$
$\lceil \log_{10} a \rceil = \left\lceil \dfrac{\ln a}{\ln 10} \right\rceil = \left\lceil \dfrac{\ln 2002^{2001}}{\ln 10} \right\rceil=\left\lceil \dfrac{2001 \ln 2002}{\ln 10} \right\rceil = \lceil 6606.2296\dots\rceil = 6697$
$$\begin{align}
\\ \lceil \log_{10} (a + b)\rceil &= \left\lceil \dfrac{\ln (a+ b)}{\ln 10} \right\rceil 
\\ & =  \left\lceil \dfrac{\ln a + \ln (1 + \dfrac{b}{a})}{\ln 10} \right\rceil 
\\ & =  \left\lceil \dfrac{\ln a + \ln \left(1 + \left(\dfrac{2}{2002}\right)^{2001}\right)}{\ln 10} \right\rceil 
\\ & =  \left\lceil \dfrac{\ln a + O(10^{-6003})}{\ln 10} \right\rceil 
\\ & =  \left\lceil \dfrac{\ln a}{\ln 10} + O(10^{-6003}) \right\rceil
\\ & =  \left\lceil 6606.2296\dots + O(10^{-6003}) \right\rceil
\\ &= 6697
\end{align}$$
