Second Loneliest Number

We change the equation as follows: $\LARGE{x^{x^{x^{16}}}=x^{16(\frac{x^{16(\frac{x^{16}}{16})}}{16})}=16}$ Then we replace $x^{16}$ with $y$ : $\LARGE{y^{(\frac{y^{(\frac{y}{16})}}{16})}=16}$

if we elevate to the power of $16$ both side of the equation then we have $\LARGE{y^{y^{(\frac{y}{16})}}=16^{16}}$ . It is trivial to see that the equation holds for $y=16.$ Thus, $\LARGE{x^{x^{x^{12}}}=x^{x^{16^{\frac{12}{16}}}}}=x^{16^{\frac{8}{16}}}=16^{\frac{4}{16}}=2$

Please consider the following infinite power tower:

$y=x^{x^{x^{x^{\dots}}}}$

$x^y=x^{x^{x^{x^{\dots}}}}=y$

$x^{x^y}=x^{x^{x^{x^{\dots}}}}=y$

$x^{x^{x^y}}=x^{x^{x^{x^{\dots}}}}=y$

Let $y=16$ and we have the following identity:

$x^{x^{x^{16}}}=x^{16}=16\Longrightarrow x=16^{\frac{1}{16}}=2^{\frac{1}{4}}$

Finding our value of $x^{x^{x^{12}}}$ :

$x^{12}=(2^{\frac{1}{4}})^{12}=2^{\frac{12}{4}}=2^3=8$

$x^{x^{12}}=(2^{\frac{1}{4}})^8=2^{\frac{8}{4}}=2^2=4$

$x^{x^{x^{12}}}=(2^{\frac{1}{4}})^4=2^{\frac{4}{4}}=2^1=\boxed{2}$

I solved the problem using the bisection method starting with initial values 1.1 and 1.2, This gives the correct answer. Here is the Matlab code.

So my method might be a little peculiar but I hope it makes sense. At first, I looked at the left side of the equation and the first thing that immediately popped into my head was 'well $2^{2 ^ {2}}$ makes up $16$ ' but that obviously did not cater to the $16^{th}$ power. So the next thought was to substitute a value in to $x$ that would make the LHS of the equation equal to $16$ . 'What if I let $x$ equal to some power of $16$ ?' And that is what I did. I substituted $16^{\frac{1}{16}}$ in place of $x$ . The LHS essentially became

$16^{\frac{1}{16}\times16^{\frac{1}{16}\times16^{\frac{1}{16}\times16}}}$

Well, great! The LHS now equals to the RHS, i.e. $16$ .

The next part was easier to calculate. Simply substitute $16^{\frac{1}{16}}$ in to the equation ${ x }^{ { x }^{ { x }^{ 12 } } }$ and solve to get RHS = $2$ .

Let $\quad y\quad ={ x }^{ 16 },\quad therefore\quad x\quad =\quad \sqrt [ 16 ]{ y } \\ Hence\quad if\quad { x }^{ { x }^{ { x }^{ 16 } } }=16,\quad { \left( \sqrt [ 16 ]{ y } \right) }^{ { \left( \sqrt [ 16 ]{ y } \right) }^{ y } }\quad =\quad 16\\ Taking\quad the\quad logarithm\quad to\quad the\quad base\quad \sqrt [ 16 ]{ y } \quad on\quad both\quad sides\quad yields\\ { \left( \sqrt [ 16 ]{ y } \right) }^{ y }{ log }_{ \sqrt [ 16 ]{ y } }\sqrt [ 16 ]{ y } \quad =\quad { log }_{ \sqrt [ 16 ]{ y } }\quad 16\quad \Rightarrow { \left( \sqrt [ 16 ]{ y } \right) }^{ y }\quad =\quad \frac { { log }_{ 16 }16 }{ { log }_{ 16 }{ \left( \sqrt [ 16 ]{ y } \right) } } \Rightarrow { \left( \sqrt [ 16 ]{ y } \right) }^{ y }\quad =\quad \frac { { log }_{ 16 }16 }{ { \frac { 1 }{ 16 } log }_{ 16 }y } \\ { \left( \sqrt [ 16 ]{ y } \right) }^{ y }\quad { log }_{ 16 }y=\quad 16,\quad which\quad means\quad that,\quad { y }^{ { \left( \sqrt [ 16 ]{ y } \right) }^{ y } }\quad =\quad { 16 }^{ 16 }\\ \\ but\quad { \left( \sqrt [ 16 ]{ y } \right) }^{ { \left( \sqrt [ 16 ]{ y } \right) }^{ y } }\quad =\quad 16,\quad implying\quad that\quad \left[ { \left( \sqrt [ 16 ]{ y } \right) }^{ { \left( \sqrt [ 16 ]{ y } \right) }^{ y } } \right] ^{ 16 }=\quad { 16 }^{ 16 }\quad =\quad { y }^{ { \left( \sqrt [ 16 ]{ y } \right) }^{ y } }\\ \quad \\ We\quad can\quad the\quad safely\quad say\quad that\\ { \left( { y }^{ { \frac { y }{ 16 } }^{ \frac { y }{ 16 } } } \right) }^{ 16 }=\quad { y }^{ { y }^{ \frac { y }{ 16 } } }\quad \Rightarrow 16\times \left( { \frac { y }{ 16 } }^{ { \frac { y }{ 16 } } } \right) ={ y }^{ \frac { y }{ 16 } },\quad \log _{ y }{ 16 } +\frac { y }{ 16 } (\log _{ y }{ y } -\log _{ y }{ 16 } )\quad =\frac { y }{ 16 } \\ Therefore,\quad (1-\frac { y }{ 16 } )\log _{ y }{ 16 } =0,\quad implying\quad that\quad the\quad answer\quad is\quad either\quad y\quad =\quad 16,\quad or\quad the\quad not\quad very\quad impractical\quad \log _{ y }{ 16 } =0\\ so\quad y\quad =\quad { x }^{ 16 }\quad =\quad 16\quad \Rightarrow x\quad =\quad \sqrt [ 16 ]{ 16 } \\$

At first I got it wrong (8) since I was not familiar with the tetration, after looking it up I still am not sure how to get the 2 algebraically, I did get it numerically, in MATLAB it's pretty simple:

sol=fsolve(@(x)x^(x^(x^(16)))-16,1) %(=1.1892)

sol_final=sol^(sol^(sol^12)) %(=2)

However you have to be carefull with the use of parenthesis, if you give matlab:

sol2=fsolve(@(x)x^x^x^16-16,1) %(=1.1421)

sol_final2=sol2^sol2^sol2^12 %(=8)

Well, I got the right answer, but I made the same dumb mistake as a couple of others did below, and just lucked into it. Is there a way to re-cast this problem so that you can't luck into the answer by just saying, "Oh, $x^{x^{x^{12}}}$ is the 4th root of $x^{x^{x^{16}}}$ , so the answer must be 2." That is clearly not a generally correct statement, since $(x^{x^{x^{12}}})^{4} = x^{4(x^{x^{12}})}$ , and $4(x^{x^{12}})=x^{x^{16}}$ ONLY in the case where $x=\pm 2^{\frac{1}{4}}$ , as in this problem.

Given x^x^x^16 = (x^x^x^12)^(x^4) =16 We now simply take the 4th root of the both sides of the equation which gives us x^x^x^12=2

X^X^X^16 = 16 and X^X^X^12 = ?

So X^X^X^12^4 = 16 and ?^4 = 16 thus ? = 16^1/4 which is 2

If x^x^x = y, then y=16^1/16 (good). Therefore y^12 = 16^(12/16) = 8 WRONG! Remember that y = x^x^x and that equals 16^(12/16), so x = 2