1. ## Logarithm numbers

If $\log_{10} (n+1) - \log_{10} n < \log_{10} 1.01$ find the smallest positive integral value for n.

2. Originally Posted by ecogreen

If $\log_{10} (n+1) - \log_{10} n < \log_{10} 1.01$ find the smallest positive integral value for n.
$\log_{10}{(n + 1)} - \log_{10}{n} < \log_{10}{1.01}$

$\log_{10}{\frac{n + 1}{n}} < \log_{10}{1.01}$

$\frac{n + 1}{n} < 1.01$

$1 + \frac{1}{n} < 1.01$

$\frac{1}{n} < 0.01$

$\frac{1}{n} < \frac{1}{100}$

$n > 100$.

Therefore the smallest possible integer value for $n$ is $101$.