Can you solve this in 1/6 minutes?

we know that $log(a) + log(b) = log(a*b), a, b > 0$ .
so, $log(\frac{5}{4}) + log(\frac{6}{5}) + ... + log(\frac{4000}{3999})$ is:

$log(\frac{5}{4} * \frac{6}{5} * \frac{7}{6} * ... * \frac{4000}{3999})$ .

Notice that $\frac{5}{4}$ and $\frac{6}{5}$ and other fractions cancel out into $\frac{4000}{4}$ .

We left with $log(\frac{4000}{4})$ or $log(1000)$ , which is $3$ .

$\large{\log_{10} \frac{5}{4} + \log_{10} \frac{6}{5} + \log_{10} \frac{7}{6} + ... + \log_{10} \frac{4000}{3999}}$

$= \large{log_{10} ( \frac{5}{4} \times \frac{6}{5} \times \frac{7}{6} \times \cdots \frac{4000}{3999} ) }$

$= \large{log_{10} \frac{4000}{4}} = \large{log_{10} 1000} = \large{\boxed{3}}$