How I Solved A NonStandard Equation

[@TheAvokela]

Replace x by 1/x. The left-hand-side of the equation becomes 2^(1/x) while the right-hand-side remains the same. We therefore have 2^x = 2^(1/x) from which it follows that x = 1/x, or, x^2 = 1.

[@ShortsOfSyber]

That’s a genius idea! 😍

[@benjaminvatovez8823]

I must disagree, I think the symmetry argument is somewhat twisted here. See this example: |x|=x. As |x|=|-x| then x=-x due to symmetry, thus x=0 would be the only solution. However, this is not correct: x=|x| if and only if x is non-negative so any non-negative real number is a solution. The symmetry argument works well in functional equation when the equation is set as true FOR ALL real x,y for instance.

[@canaDavid1]

2^x - x - 1/x has a derivative ln(2)2^x - 1 + 1/x² which is positive for all x > 0, hence it has at most one solution there, and x=1 trivially works. For x < 0, 2^x > 0 > x + 1/x.

[@fadetoblah2883]

Not sure that the method I used was sound, but it got me (perhaps accidentally) to the correct conclusion. Here goes: 2ˣ = x + 1/x x² - (2ˣ)x + 1 = 0 The quadratic formula gives us that x = [2ˣ ± √(2²ˣ - 4)] / 2 It seems to me that this equation can only make sense if it yields a single value of x (otherwise you enter a weird realm where one value of x on the right side yields two values of x on the left), which implies that the discriminant should be equal to 0. Therefore 2²ˣ - 4 = 0 2²ˣ = 4 x = 1, and this should really be the only solution.

[@MichaelRothwell1]

From your quadratic you can deduce that the discriminant Δ=2²ˣ-4≥0, so x≥1. For x=1, we get a solution. For x>1, you get that x must satisfy one of two equations (note that x need only satisfy one of these equations, not both), so you still need to show that there is no solution to either equation for x>1.

[@MichaelRothwell1]

Nice problem! Here's my solution. For x0 and x+1/x0, x+1/x=(x²+1)/x=((x-1)²+2x)/x =(x-1)²/x+2≥2, with equality only when x=1. For 01, x+1/x

[@elmer6123]

Why not just plot y=2^x and y=x+1/x and see that the two curves intersect only at the point x=1 and y=2? Am I missing something?

[@FisicTrapella]

That's what he did. I missed a more "mathematical" method....