<s>[INST] Summing reciprocal logs of different bases I recently took a math test that had the following problem:
$$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
The sum is equal to 1. I understand that the logs can be broken down into (first fraction shown)
$$
\frac{1}{\log_{2}1 + \log_{2}2 + \log_{2}3 + \dots + \log_{2}50}
$$
How do the fractions with such irrational values become $1$? Is there a formula or does one simply need to combine fractions and use the basic properties of logs?
 [/INST] $$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
$$=\log_{50!}2+\log_{50!}3+\log_{50!}4+...+\log_{50!}50$$
$$=\log_{50!}(2\cdot3\cdot4\cdot...50)$$
$$=\log_{50!}(50!)$$
$$=1$$
</s>