Split A Log

$\log 10 + \log 20 > 2\log N$

Using the rule: $\log_{b}a + \log_{b}c = \log_{b}ac$

$\log 200 > 2 \log N$

Using the rule: $a \log b = \log b^{a}$

$\log 200 > \log N^{2}$

Then, you can drop the logarithms because if $\log_{b}a = \log_{b}c$ , then $a = c$ .

$200 = N^{2}$

Taking the square root of $200$ is $10\sqrt{2}$ , which is approximately $14.14$ . Therefore, the largest $N$ is $\boxed{14}$ .

Honestly a base value should be given just for the sake of it