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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:

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

Bongo Man
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    What you tried is not valid. Generally speaking, $(a+b+c)^p \neq a^p + b^p + c^p$ (the "freshman's dream"). – MPW Aug 08 '23 at 13:03
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    You know that $(a+b)^x$ isn't the same as $a^x+b^x$, right? – Gerry Myerson Aug 08 '23 at 13:04
  • My goal is to create an equation taking $2^x$ as some variable, say $a$, then solve for $x$ with the solution of that equation with $a$. For example, if the eqn. was $2^{2x} - 2^{x+1} + 12 = 0$, then taking $2^x$ as $a$, we would get $a^2 - 2a + 12 = 0$ – Bongo Man Aug 08 '23 at 13:09
  • I don't think this equation can be solved in that way. By the way, it can be verified that $x=3$ is the unique real solution of the equation. – Sangchul Lee Aug 08 '23 at 13:10
  • @SangchulLee you are correct, but how do we derive that $x = 3$? – Bongo Man Aug 08 '23 at 13:13
  • If you have no fancy way of reducing an equation into a simpler one, what remains is a trial and error. And that's what I did. It is a bit of a bummer, but I would be surprised if there is any better approach. – Sangchul Lee Aug 08 '23 at 13:15
  • Anyway, here is my thought process that led me to the answer: If the problem was designed to be solvable by hand, then it probably has a rational solution, or even an integer solution. Since it is hard to cancel out irrational numbers by subtraction, a root of the equation would probably make the term $2^{6/x}$ a rational number as well. So tried $x=1,2,3,6$, and $x=3$ worked. – Sangchul Lee Aug 08 '23 at 13:30
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    Or you could notice that 2^4-2^2=12, and arrive at x=3 slightly more systematically! – avighnac Aug 08 '23 at 13:40
  • My guess is that the point of the question is to prove that the root is unique. It's not hard to find $x=3$, it's more interesting to look at whether that's the only root. – Chris Lewis Aug 08 '23 at 13:46

2 Answers2

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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 $x=3$.

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

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.

IV_
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