the end-of-year exponential equation!

[Lastrevio]

You also have to prove that the equation has an unique solution. You do that by stating that the left hand side is an exponential function (thus strictly increasing), and the right hand side is a constant, and the intersection between the graph of a strictly increasing function and a constant function is always one single point.

[Iocun]

Great problem! My line of thought:

x^x^x^2017=2017 --> raise x to the power of each side -->
x^x^x^x^2017=x^2017
let a = x^2017 -->
x^x^x^a = a --> ahah! this looks like what we had originally. a must be 2017.
a = x^2017 = 2017
x = 2017^(1/2017)

[natehoffmaster]

You know things are going to get intense when he pulls out the blue pen.

[frostcrackle]

2017 was a really good year,

*ｉｓｎ＇ｔ ｉｔ？*

[violetemmott]

I did this my way, I turned it into an infinite sum x^x^x^x^x^x^x... =2017 and then simplified it to x^2017=2017
And got the 2017th root of 2017 as my answer

[carpetjuise]

Me : *looks at thumbnail*
My stupid ass thinking Im smart : *X is obviously 1*

[thephysicistcuber]

Amazing solution!

I came up with something else:

Let y=2017

x^x^x^y=y
-> x^x^x^(x^x^x^y)=y
-> x^x^x^( x^x^x^(x^x^x^y))=y
...
-> Infinite tower of x^x...=y
-> x^(infinite tower)=y
-> x^y=y
-> x=y^(1/y)

[SeaYou]

My mans flexing with the Supreme jacket 😂😂😂👌

[ffggddss]

"Isn't it?" is #1.
#2 is "... for you guys."

[deeptochatterjee]

You can start from the easier equation and then raise x to the power of each. On the left you get the intermediately difficult equation and on the right you get (2017^(1/2017))^2017 Wich gets you back to 2017. From there you can conclude that no matter how many X's there are your original solution for x will always be an answer.

[atuuuuum]

so if we have x^x^x^x^n=n the answer will be n^(1/n)?

[badrunna-im]

for x^x^x^x^... ad infinitum=2017, following the exponential order of operations, you can let y=x^x^x^...=2017, such that x^y=2017 or x^2017=2017, which reduces it to the first case. So in a way, it does work for "infinity". (Idea from Mind Your Decisions' Infinite Exponent Tower video.)

[gadxxxx]

I look forward to many, many more great integrations etc in 2018. Have a great year.

[ethanalaniz]

I honestly love using logic as a way of solving complex equations

[ffggddss]

After considerable poking around on a calculator, I started thinking along the lines of solving one of those infinite power tower equations, where you just say, x^... = a ; xª = a ; x = ª√a .
Namely, I finally realized that if x²⁰¹⁷ = 2017, then that value will "propagate" from the top, all the way down.
So the solution is x = ²⁰¹⁷√2017 =

Because, given x²⁰¹⁷ = 2017, the 'tower' collapses, thus:

x^(x^(x^2017)) = x^(x^2017) = x^2017 = 2017

[sonaruo]

log_x (X^X^X^2017)= log_x(2017) we log both sides with log base x
X^X^2017= log_x(2017) we simplify the first part
X^X^2017=x^2017 we solve for second part
X^2017=2017 base are the same so much also be the powers
and you are done

[analyticssingh]

I was thinking like:-
X^x^x^2017=2017
But 2017 on left is x^x^x^2017
If we do repeated substitution for 2017 on left
We get,

x^2017=2017
x=2017^1/2017

[mateuszokulus]

Other solution. Plug left hand side in place of 2017 on and on, acquiring an infinite power tower = 2017. Infinite tower is x^(infinite towerl So x^2017=2017, and we find a root and we are done.

[hhhhhh]

Take log_x on both sides 3 times
2017 = log_x(log_x(log_x(2017)))
Repeatedly applying log_x can't go down and back up or up and back down; it has to be a fixed point
So, log_x(2017) = 2017, which can be rearranged to x^2017 = 2017, so x = 2017 ^ (1 / 2017)

[danielfernandocarballo]

And similarly, for all value a
a > 0,
X ^ X ^ X ^ a = a
implies that
X = a ^ (1 / a)