So many logs and exponents

Calculus Level 4

J = n + 1 e n + 2 e d x x i = 1 n ln ( ln ( ln ( ln x ) ) ) i times \Large J=\int_{^{n+1}e}^{^{n+2}e} \dfrac {dx}{ x \prod_{i=1}^n \underbrace{\ln ( \ln ( \ln \cdots ( \ln x ) ) \cdots )}_{i \text{ times} }}

Let A B ^A B denote the tetration function, A B = B B B B A times ^A B = \underbrace{B^{B^{B^{\cdot^{\cdot^{\cdot^B}}}}}}_{ \text{A times} } .

Find J + 2 \lfloor J + 2 \rfloor .


The answer is 3.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

L i ( x ) = ln ( ln ( ln ( ln x ) ) ) i times L_{i}(x) = \underbrace{\ln ( \ln ( \ln \cdots ( \ln x ) ))}_{i \text{ times}}
J = n + 1 e n + 2 e 1 x L 1 ( x ) L 2 ( x ) L n ( x ) d x J = \displaystyle \int_{ ^{n+1}e}^{ ^{n+2}e} \dfrac{1}{xL_{1}(x)L_{2}(x)\ldots L_{n}(x)} dx
L n ( x ) = t L_{n}(x) = t
1 x L 1 ( x ) L 2 ( x ) L n 1 ( x ) d x = d t \dfrac{1}{xL_{1}(x)L_{2}(x) \ldots L_{n-1}(x)}dx = dt
J = e e e d t t = [ ln ( t ) ] e e e = e 1 \therefore J =\displaystyle \int_{e}^{e^{e}} \dfrac{dt}{t} = \left[ \ln(t) \right]_{e}^{e^{e}} = e - 1
J + 2 = e 1 + 2 = e + 1 \lfloor J + 2 \rfloor = \lfloor e - 1 + 2 \rfloor = \lfloor e + 1 \rfloor


2 < e < 3 2 < e < 3
3 < e + 1 < 4 3 < e+ 1 < 4
J = e + 1 = 3 J = \lfloor e + 1 \rfloor = 3

Great solution +1

Prince Loomba - 4 years, 11 months ago

Cant believe its level 4

Prince Loomba - 4 years, 9 months ago
Prince Loomba
Jul 5, 2016

It is greatest integer of j not j!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...