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Is there a simple way of proving that x=2^x can't happen for real x? Without series expansion.

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Is there a simple way of proving that x=2^x can't happen for real x? Without series expansion.

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arildno

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Sure!

1. Note that if there were a solution, x cannot be negative, since the right-hand side is positve.

2. Note that at x=0, we have 2^x=1.

3. In order for "x" then, to become equal, (or greater) to 2^x, the slope of the function f(x)=x must be greater than the slope of the function g(x)=2^x prior to the x that makes f=g

4. Now, the slope of f is equal to 1, for g, it equals 2^xln(2)>0

5. Now, we see that g is strictly increasing.

IF the graph of f is to intersect somewhere with the graph of g, then the point at which g has equal slope as f must either be a point of intersection, OR IN A REGION IN WHICH f is greater than g!

(Since the graph of g will eventually undertake the graph of f, and if the point where g's slope is equal to f's slope is prior to f's intersection with g, then f will NEVER intersect with g)

6. Thus, we find that when g's slope equals f's slope, this happens at X=ln(1/ln(2))

7. At that point, we have f(X)=ln(1/ln(2)), whereas g(X)=(1/ln(2))^ln(2)

8 Since g(X)>f(X) we may conclude that g is always greater than f.

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Much simpler is to write 2^x - x = 0, and take the derivative of LHS: ln(2)2^x - 1. On [0, ∞), this is clearly strictly positive, so the function 2^x - x is strictly increasing on that interval. At x = 0, 2^x - x = 1, so it never attains 0 on [0, ∞) (it only gets larger than 1). It also can never attain 0 on (-∞, 0), so there is no solution.

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arildno

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Most definitely not!Much simpler is to write 2^x - x = 0, and take the derivative of LHS: ln(2)2^x - 1. On [0, ∞), this is clearly strictly positive,

At x=0, for example, ln(2)-1 is negative.

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Ah, then I retract my solution.

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Ah, thanks for the replies!

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uart

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This question could be turned into a nice little exercise for introductory calculus.

Q. Find the largest value of "

a = e^(1/e), with the solution occurring at x = e.

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