Thanks a lot for all your help in advance.
If $\displaystyle \log_{10} (n+1) - \log_{10} n < \log_{10} 1.01 $ find the smallest positive integral value for n.
$\displaystyle \log_{10}{(n + 1)} - \log_{10}{n} < \log_{10}{1.01}$
$\displaystyle \log_{10}{\frac{n + 1}{n}} < \log_{10}{1.01}$
$\displaystyle \frac{n + 1}{n} < 1.01$
$\displaystyle 1 + \frac{1}{n} < 1.01$
$\displaystyle \frac{1}{n} < 0.01$
$\displaystyle \frac{1}{n} < \frac{1}{100}$
$\displaystyle n > 100$.
Therefore the smallest possible integer value for $\displaystyle n$ is $\displaystyle 101$.