# 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?

• In this case you will see that the properties of log will give you your answer – R_D Oct 28 '15 at 14:28

$$\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}$$ $$=\log_{50!}2+\log_{50!}3+\log_{50!}4+...+\log_{50!}50$$ $$=\log_{50!}(2\cdot3\cdot4\cdot...50)$$ $$=\log_{50!}(50!)$$ $$=1$$

• This uses a rule I'd never heard of! rapidtables.com/math/algebra/Logarithm.htm#base switch – Dancrumb Oct 28 '15 at 16:25
• @Dancrumb Which rule are you talking of? – SchrodingersCat Oct 28 '15 at 16:32
• The one I linked to – Dancrumb Oct 28 '15 at 17:16
• In case the link doesn't work: The logarithm base switch rule. – Dancrumb Oct 28 '15 at 17:17
• Presumably $\log_a( b) = 1/ \log_b( a)$ which can be shown from $b^{\log_b(a) \log_a(b)} = \left(b^{\log_b(a)}\right)^{\log_a(b)} = a^{log_a(b)} = b$ so $\log_b(a) \log_a(b) = 1$ – Henry Oct 28 '15 at 21:01

HINT: use that $\log_2 50!=\frac{\ln(50!)}{\ln(2)}$ thus we get $$\frac{\ln(2)+\ln(3)+...+\ln(49)+\ln(50)}{\ln(50!)}=\frac{\ln(50!)}{\ln(50!)}=1$$

• Not complaining, but why did you call it a hint, when you gave the entire solution? – Paul Sinclair Oct 29 '15 at 0:49

$$\sum_{k=2}^{50}\frac{1}{\log_k(50!)}=\sum_{k=2}^{50}\frac{1}{\frac{\log(50!)}{\log k}}=\sum_{k=2}^{50}\frac{\log k}{\log(50!)}=\frac{1}{\log(50!)}\sum_{k=2}^{50}\log k=\frac{\log(50!)}{\log(50!)}.$$

This is a general result. We can write for any $N$

\begin{align} \sum_{n=2}^N\frac{1}{\sum_{m=2}^N \log_n(m)}&=\sum_{n=2}^N\frac{1}{\sum_{m=2}^N \frac{\log_b (m)}{\log_b(n)}}\\\\ &=\frac{\sum_{n=2}^N\log_b(n)}{\sum_{m=2}^N\log_b(m)}\\\\ &=1 \end{align}

where we used $\log_n(m)=\frac{\log_b(n)}{\log_b(m)}$. And we are done!