# Evaluating $\frac{1}{\log_2 (100!)} + \cdots + \frac{1}{\log_{100}(100!)}$

Firstly, I am completely aware that logarithms can be rewritten in exponential form. So, I do know at least something about logs. :)

However, given the task to simplify and rewrite the expression, $${\frac1{\log_2(100!)} + \frac1{\log_3(100!)} + \frac1{\log_4(100!)} + ... + \frac1{\log_{99}(100!)} + \frac1{\log_{100}(100!)} }$$

Without any logarithms, I'm a little stumped.

Before I give my guesses as to how to solve it, I am a little hazy on my algebra sometimes, especially logs. I do know that,

$${\log_b(x) = y}$$ and $${x = b^y}$$

are just inverses of each other. This is my first clue as to how to eliminate the logs from this. My uneducated guess would be

$${\log_{n+1}(100!) = y}$$ and $${(100!) = {}(n+1)^y}$$

From there, if that's even remotely correct, I can only guess that there is something dead obvious I'm missing. Without knowing that, sigma notation is where I'd start. Like,

$${\sum_{i = 1}^{99} \frac1{\log_{n+1}(100!)} }$$

And then replacing the denominator with the inverse. I think?

$${\sum_{i = 1}^{99} \frac1{{(100!) = {}(n+1)^y}} }$$

But then there's that = sign in the denominator which is no good at all! I still don't know what y equals!

That's about how lost I am.

I'm grateful for any measurement of enlightenment!

Cheers,

-Jon

When you've got a sum of fractions, the easiest thing to do is to have all the denominators be the same. This problem would also be a lot easier if all the bases of the logarithms were the same. So we use the change of base rule: $\log_a b = \frac{\log_c b}{\log_c a}$ for any other base $c\neq 1$. So your sum of fractions becomes: $$\frac{1}{\log_2(100)!} + \cdots + \frac{1}{\log_{100} (100!)} = \frac{1}{\frac{\log (100!)}{\log 2}} + \cdots + \frac{1}{\frac{\log (100!)}{\log 100}} = \frac{1}{\log(100!)}(\log 2 + \log 3 + \cdots + \log 100)$$

Now remember that $\log a + \log b = \log ab$ so the expression above simplifies to the very neat $$\frac{\log (2\cdot 3 \cdots 100)}{\log 100!} = \frac{\log 100!}{\log 100!} = 1$$

NB: I've omitted the base of the logarithm because it becomes irrelevant what base you choose, you can pick whichever one you want or interpret $\log$ as $\log_e$.

• Thanks so much for this! My textbook is only as clear as those who understand the jargon; to me, it's as clear as mud. Many thanks. :) – Werewoof Apr 20 '17 at 20:43
• @Werewoof Thanks very much the feedback, I'm glad you understood everything! – Zain Patel Apr 20 '17 at 21:04
• I insist, thank YOU! :) Very helpful information. – Werewoof Apr 21 '17 at 15:30

Here's a hint: $log_ab = \frac{\log{b}}{\log{a}}$. Apply this to each of the denominators, and the rest should be smooth sailing.

• Oooooh, I see how. :) Awesome, that makes sense in conjunction with the answers. – Werewoof Apr 20 '17 at 20:44

The key is that $$\log_a{b} = \frac{\log_c{b}}{\log_c{a}},$$ where $c \neq 1$ is some other base. The other important property is $$\log_c{(ab)} = \log_c{a}+\log_c{b}$$ for any positive $a$ and $b$ ($c^{A}c^B=c^{A+B}$).So $$\frac{1}{\log_n{(100!)}} = \frac{\log_c{n}}{\log_c{(100!)}},$$ and then $\sum_{n=2}^{100} \log_c{n} = \log_c{(100!)}$. Hence the expression is just $1$.

• I think I follow, thanks! :) – Werewoof Apr 20 '17 at 20:48
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