Is there any way express $2^{6/x} - 2^{x+1} + 12 = 0$ taking $2^x$ as some variable, say $a$?

Question

I tried raising to $x$ on both sides, getting $2^6 - 2^{x(x+1)} + 12^x = 0$, but we still can't simplify the $2^{x(x+1)}$ in terms of $2^x$. Is there any method of simplifying the equation, taking $2^x$ as some variable, say $a$?

Edit: I got this equation by simplifying the following expression:

$$\sqrt[x]{64} - \sqrt[3]{2^{3x + 3}} + 12 = 0$$

Answer 1

As others have mentioned, the idea of substituting $a=2^x$ won't get you very far with this question, because the $2^\frac{6}{x}$ term will still cause problems.

It's not strictly true to say that trial and error is the only way to proceed, though. We can be more rigorous than just observing that $3$ is a root; we can show that it's the only real root of the equation. I suspect this is what whoever asked the question is after.

Rewrite the equation as $f(x)=g(x)$ where $$f(x)=2^\frac{6}{x},\quad g(x)=2^{x+1}-12$$

$g(x)$ is clearly a strictly increasing function in $x$, and is defined for all real $x$.

$f(x)$ is not defined at $x=0$, but is everywhere else.

$f(x)>0$ for all $x\neq0$; the $g(x)<0$ for all $x<\log_2 {6}$, so we can restrict ourselves to looking for roots where $x\ge1$. Note that $f(1)>g(1)$.

It's fairly easy to see that the LHS is a strictly decreasing function for $x>1$ (direct from the definition) so the two curves $$y=f(x);\quad y=g(x)$$

intersect in exactly one point $x>1$.

So, if we find a point where the two functions are equal, it is that unique intersection point.

If the numbers weren't so nice (if, eg, we replaced the $12$ in the question with an $11$), we'd need to use a numerical method (eg Newton's method, or bisection), but a sketch of the curves and a simple table of values quickly finds the unique root $x=3$.

Answer 2

$$2^\frac{6}{x}-2^{x+1}+12=0$$ $$(2^\frac{1}{x})^6-2\ 2^x+12=0$$

E.g. with help of Lindemann–Weierstrass theorem, we find that the terms $2^\frac{1}{x},2^x$ are algebraically independent. Therefore, they cannot both fulfill the same algebraic equation simultaneously. $2^\frac{1}{x}$ cannot be expressed as an algebraic function of $2^x$ and vice versa therefore.

We see, your equation is a polynomial equation of more than one algebraically independent monomials ($2^\frac{1}{x},2^x$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

It seems your equation can be solved neither with Lambert W nor with Generalized Lambert W or Hyper Lambert W.