Ever Seen Nested Logarithms?

Calculus Level 2

log ⁡ log ⁡ log ⁡ log ⁡ … 27 27 27 27 = ? \LARGE \log_{\log_{\log_{\log_{\ldots} 27 } 27 } 27 } 27 = \ ?

Details and Assumptions :

Note that the base of the logarithm is nested indefinitely.


The answer is 3.

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Seth Lovelace by Seth Lovelace , 23, USA

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4 solutions

We have, l o g l o g l o g . . . 27 27 27 = L \displaystyle log_{log_{log_{...27}}27}27 = L

Hence, l o g L 27 = L \displaystyle log_L 27 =L

Raising both sides to L, we get : L l o g L 27 = L L \displaystyle L^{log_L 27} = L^L

Since, 3 3 = L L ⇒ L = 3 \displaystyle 3^3 = L^L \Rightarrow \boxed{L=3}

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Moderator note:

This solution is incomplete. It shows that if a numerical limit exists, then the answer must be 3. However, it has not shown a limit must exist.

Don't you have to show that the nested logarithm (defined as a limit) actually exists? You show that if it exists, then it is 3... If you think I'm nitpicky, what would the answer be if we were to replace 27 by 4 in the problem?

Otto Bretscher - 6 years, 2 months ago

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Thank you sir, for pointing that out! Helped me realise that the simplest of problems have the deepest of consequences :)

B.S.Bharath Sai Guhan - 6 years, 2 months ago

as L^L=4 => L=2

Somesh Singh - 6 years, 2 months ago

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But that's not the answer! In fact, in that case the limit fails to exist... That was my point: the given solution is incomplete.

Otto Bretscher - 6 years, 2 months ago

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@Otto Bretscher – Thanks for highlighting that. I've added a note to the solution. I've also created a problem to highlight this - Inspired by Otto Bretscher .

It is common for such nested equations that people do not consider if the limit actually exists.

Calvin Lin Staff - 6 years, 2 months ago

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@Calvin Lin – Thank you for highlighting this important issue, Calvin! It can be very hard to show that such a "nested limit" actually exists... consider the case of the power tower of i you posed the other day. Nobody proved the existence of that limit (including myself)... we just found the fix point of the iteration and hoped for the best (maybe based on some numerical experimentation).

Otto Bretscher - 6 years, 1 month ago

@Calvin Lin – You are right sir, I did assume the existence of said limit, and I was pleasantly surprised by Mr.Bretscher's instance! The question now is, how would I rigorously prove that said limit exists?

B.S.Bharath Sai Guhan - 6 years, 2 months ago

Sir how do we show that the limit actually exists or does not exist in such nested function??

Puneet Pinku - 5 years, 1 month ago

How are we supposed to solve if L^L=10 ??

WolframAlpha

Santanu Banerjee - 6 years, 5 months ago

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Well Santanu, I see that you are a level 4 in Calculus, so one method that would work to get the solution of x x { x }^{ x } = 10 would be to use the Newton-Raphson root approximation method, which is an algorithm that successively gives better and better approximations for the root you are seeking by using derivatives to "hone in" on where the function intersects the x-axis. For example, x x { x }^{ x } = 10 can be written as x x − 10 { x }^{ x }-10 = 0 , which we can use as a function, and then find the root of the function. Another method that would work (without calculus) is the bisection method, which is essentially where you take a function f(x), determine that a root falls between some a and b (as one side must be negative and the other positive, unless the root has an even multiplicity), and continually narrow the interval by adding a+b and then dividing by 2. By redefining the interval that the root falls between, either [a,(a+b)/2] or [(a+b)/2,b], one can repeat to generate successively more accurate approximations of the root in mind.

Seth Lovelace - 6 years, 5 months ago

I think we can estimate it. For example, if x x = 10 x^x=10 , take log both sides (with base 10 10 )

x log ⁡ x = 1 x\log x=1

We know it lies between 2 and 3, Let x = 2.5 x=2.5 , x log ⁡ x = 2.5 log ⁡ 2.5 = 2.5 × 0.3980 ≈ 2.5 × 0.4 = 1 x\log x=2.5\log 2.5\\=2.5×0.3980≈2.5×0.4=1

So, x ≈ 2.5 x≈2.5

Luckily, I got it on first chance, else I must have tried with x=2.7 (as I know value of log ⁡ 2.7 \log 2.7 ) or x=2.4 or 2.25.

Pranjal Jain - 6 years, 5 months ago

Nice solution, did by the same way.

Anuj Shikarkhane - 6 years, 5 months ago

I don't get why log L 27=L?

Katherine Hong - 6 years, 1 month ago

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It's because if you cover up the first log and first 27 with your hand, you can you'll see its the same thing again. This only happens when things are infinitely nested

Shaumik Khanna - 3 years, 6 months ago
Govind Balaji
Jan 8, 2015

The above problem can be written as log ⁡ L 27 = L \log_L 27=L By the definition of logarithm, this can also be written as L L = 27 = 3 3 L^L=27=3^3

Thus L = 3 L=3

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Chew-Seong Cheong
Apr 12, 2015

Let log ⁡ log ⁡ log ⁡ . . . 27 27 27 = x = log ⁡ x 27 ⇒ x x = 27 ⇒ x = 3 \quad \log_{\log_{\log_{...} {27}}{27}} {27} = x = \log_x {27} \quad \Rightarrow x^x = 27 \quad \Rightarrow x = \boxed{3}

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Firstly, L>0 and L ≠ 1 L \neq 1 (obvious). We can see that l o g L 27 = L log_L 27=L

→ l n 27 l n L = L → 3. l n 3 = L . l n L \rightarrow \frac{ln 27}{ln L}=L \rightarrow 3.ln3=L.lnL

The function f ( x ) = x . l n x f(x)=x.lnx increase when x in ( 1 ; + ∞ ) (1;+\infty) and is negative when x in ( 0 ; 1 ) (0;1) (proven using calculus), so x=3 is the only root.

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