Evaluating several powers of ##x## from a given equation

[brotherbobby]

Homework Statement

If $\quad\boldsymbol{\displaystyle x^{\displaystyle 2x^{\displaystyle 6}}=3}$, evaluate $\quad\boldsymbol{\left(x^{\displaystyle x^{\displaystyle x^{\displaystyle 6}}}\right)^{\displaystyle\sqrt 3} = ?}$

Relevant Equations

1. $a^m \cdot a^n = a^{m\cdot n}$
2. $\left(a^m\right)^n = a^{mn}$.
3. If $a^x=b\Rightarrow x\log a=\log b\quad \text{to any base}$

(I can't think of anything else given the little I was able to do in my way of an $\text{attempt}$.)

Problem statement : If $\boldsymbol{\displaystyle x^{\displaystyle 2x^{\displaystyle 6}}=3}$, evaluate $\boldsymbol{\left(x^{\displaystyle x^{\displaystyle x^{\displaystyle 6}}}\right)^{\displaystyle\sqrt 3} = ?}$.

Attempt : I copy and paste my attempt using Autodesk Sketchbook$^{\circledR}$. I hope I am not violating anything.

Doubt : If only I could evaluate $x$, or so it would seem. Note I have an irrational power $^{\sqrt 3}$ to negotiate with as well.

A help or a hint would be welcome.

 

[PeroK]

You need to be carfeul. $$x^{x^{x^6}} = x^{(x^{x^6})} = x^{(x^{(x^6)})}$$

 

[anuttarasammyak]

Homework Statement:: If $\quad\boldsymbol{\displaystyle x^{\displaystyle 2x^{\displaystyle 6}}=3}$, evaluate $\quad\boldsymbol{\left(x^{\displaystyle x^{\displaystyle x^{\displaystyle 6}}}\right)^{\displaystyle\sqrt 3} = ?}$

I observe
x=3^{\frac{1}{6}}
satisfies "If ...,".

Say
x^6=y
y^{2y/6}=3
y\ln y=3\ln 3
y=3
x=3^{\frac{1}{6}}

 

[DrClaude]

I observe
x=3^{\frac{1}{6}}
satisfies "If ...,".

If $x^{(2(x^6))} = 3$ but not if $x^{((2x)^6)} = 3$. The problem statement is ambiguous.

 

[PeroK]

If $x^{(2(x^6))} = 3$ but not if $x^{((2x)^6)} = 3$. The problem statement is ambiguous.

It's not much different from $e^{-\frac 1 2 x^2}$, which is unambiguous.

 

[DrClaude]

It's not much different from $e^{-\frac 1 2 x^2}$, which is unambiguous.

The thing is I read the problem as $x^{(2(x^6))} = 3$ (as in your example) but the OP as $x^{((2x)^6)} = 3$. So which is it?

 

[PeroK]

The thing is I read the problem as $x^{(2(x^6))} = 3$ (as in your example) but the OP as $x^{((2x)^6)} = 3$. So which is it?

The convention is definitely the former. $e^{kx^2}$ being the example to bear in mind.

The OP has misinterpreted both expressions, I think.