What the log?

We can simplify the given expression by using change of base theorem which goes like this : $log_{p}{q} = \frac{log_{k}{q}}{log_{k}{p}}$

The expression will get simplified to , $\lim_{x \to 1} log(x+1)\times (\frac{log(x+2)}{log(x+1)}) \times (\frac{log(x+3)}{log(x+2)}) .......\times (\frac{log(x+98)}{log(x+97)}) \times (\frac{log(x+99)}{log(x+98)})$

which simplifies to, $\lim_{x \to 1} log(x+99)$

$\Rightarrow log 100 = \boxed {2}$

So we get the answer as $2$