Logs with Factorials!

Algebra Level 1

1 log 2 ( n ! ) + 1 log 3 ( n ! ) + 1 log 4 ( n ! ) + + 1 log n ( n ! ) = ? \frac{1}{\log _{2} {(n!)}} + \frac{1}{\log _{3} {(n!)}} + \frac{1}{\log _{4} {(n!)}} + \ldots + \frac{1}{\log _{n} {(n!)}} = ?


The answer is 1.

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2 solutions

Mahdi Raza
May 6, 2020

Animated video solution: can be found here

Property References:

1. \color{#3D99F6}{1.} log ( a b ) = log ( a ) + log ( b ) 2. \log {(ab)} = \log {(a)} + \log {(b)} \\ \color{#3D99F6}{2.} log a b = 1 log b a \log_{a} {b} = \frac{1}{\log_{b} {a}}

Evaluating the expression

= 1 log 2 ( n ! ) + 1 log 3 ( n ! ) + 1 log 4 ( n ! ) + + 1 log n ( n ! ) = log n ! ( 2 ) + log n ! ( 3 ) + log n ! ( 4 ) + + log n ! ( n ) [ 2 ] = log n ! ( 2 3 4 5 n ) [ 1 ] = log n ! ( n ! ) = 1 \begin{aligned} &= \frac{1}{\log _{2} {(n!)}} + \frac{1}{\log _{3} {(n!)}} + \frac{1}{\log _{4} {(n!)}} + \ldots + \frac{1}{\log _{n} {(n!)}} \\ \\ &= {\log _{n!} {(2)}} + {\log _{n!} {(3)}} + {\log _{n!} {(4)}} + \ldots + {\log _{n!} {(n)}} \quad [{\color{#3D99F6}{2}}] \\ \\ &= {\log _{n!} {(2 \cdot 3 \cdot 4 \cdot 5 \cdots n)}} \quad [{\color{#3D99F6}{1}}] \\ \\ &= {\log _{n!} {(n!)}} \\ \\ &= \boxed{1} \end{aligned}

David Vreken
May 6, 2020

1 log 2 ( n ! ) + 1 log 3 ( n ! ) + 1 log 4 ( n ! ) + . . . + 1 log n ( n ! ) \displaystyle \frac{1}{\log_2 (n!)} + \frac{1}{\log_3 (n!)} + \frac{1}{\log_4 (n!)} + ... + \frac{1}{\log_n (n!)}

= 1 log ( n ! ) log 2 + 1 log ( n ! ) log 3 + 1 log ( n ! ) log 4 + . . . + 1 log ( n ! ) log n = \displaystyle \frac{1}{\frac{\log (n!)}{\log 2}} + \frac{1}{\frac{\log (n!)}{\log 3}} + \frac{1}{\frac{\log (n!)}{\log 4}} + ... + \frac{1}{\frac{\log (n!)}{\log n}}

= log 2 log n ! + log 3 log n ! + log 4 log n ! + . . . + log n log n ! = \displaystyle \frac{\log 2}{\log n!} + \frac{\log 3}{\log n!} + \frac{\log 4}{\log n!} + ... + \frac{\log n}{\log n!}

= log 2 + log 3 + log 4 + . . . + log n log n ! = \displaystyle \frac{\log 2 + \log 3 + \log 4 + ... + \log n}{\log n!}

= log 2 3 4 . . . n log n ! = \displaystyle \frac{\log 2 \cdot 3 \cdot 4 \cdot ... \cdot n}{\log n!}

= log n ! log n ! = \displaystyle \frac{\log n!}{\log n!}

= 1 = \displaystyle \boxed{1}

Great solution @David Vreken !

Mahdi Raza - 1 year, 1 month ago

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Thank you! I like yours better as it was a little bit quicker, but I wanted to show a slightly different way.

David Vreken - 1 year, 1 month ago

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Thanks!, this problem is actually inspired by 3b1b's lockdown math ep. 6. He shows a similar question and solves just like your method. Here is the link

Mahdi Raza - 1 year, 1 month ago

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@Mahdi Raza Thanks for the link!

David Vreken - 1 year, 1 month ago

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