Solving the Exponential Equation x^x=2^{1-x^2}

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[laman]

What I see a lot on Y'T regarding math is that almost all of the math problems are how to solve them. Personally, I think the bigger challenge, and what I certainly would like to learn is, how to model a real life problem into a math expression )function or formula). Take for example the right hand side of this problem. What are the steps or tools to evaluate a real life problem and how to discover that the behavior of that problem has the mathematical form as expressed on the right hand side. I think that is more rewarding. There is so much on solving and barely anything on how to model a problem into a math expression. Moreover, math software has become so sophisticated now that solving is done virtually. But nonetheless, I really enjoy the way you solve equations and provide the basic math theory and graphs to better understand the problem. I wonder if any suggestions can be made regarding the issue I raised.

[SuperYoonHo]

Thank you so much sir!!! These videos mean a LOT to me, they are so helpful!!!😍😍

[ismaelperbech]

Solving this equations can be reduced by looking for invariance. If you compose (1-x^2) o (-x) you get 1-x^2 and in this way the equation turns out to be x^x = (-x)^(-x) which is way easier.

[mkwfqem]

We know, x > or = 0. Can we use monotony, then x >=1, because then x € [0;1], y = x^x < or = 1, but 2^(1-x²) diminish, and the x €[0;1], y > or = 1. And in this interval we haven't any roots.
Then x > or = 1 : x^x grow and 2^(1-x²) diminish. And we can say: we have no more than 1 root. We can find root. X = 1

[dyip-vbwl]

Pardon my lack of mathematical chops, but at 4:58, how exactly did you figure out that f’ from the right is positive and left is negative? I only knew after graphing the derivative of the function

[peterchan]

1:45
Please elaborate on why you could justify setting them to 0.

[christianthomas]

for solving this equation I've used log base 2 so here every time log is written it's (log base 2)
logging the two members I get:log(2^(x^2).x^x) = log(2)= 1 so: log(2^(x^2)) log(x^x) = 1 and as log(2)=1, I get the simplified equation:
x^2 + x.log(x) = 1. As obviously x must be greater than 0, and greater than 1 (because 2^(log(x)) = 2^(1-x^2) ) therefore : x^2 + x.log(x) IS GREATER OR EQUAL to 0, WHICH MEANS the only solution is 1 to have: 1^2 + 1.log(1) = 1 .

[MrLidless]

x^x *is* defined for negative numbers, being a complex number when x is not a negative integer, and a real number otherwise.

[MichaelRothwell]

This nice equation can easily be solved for x>0 without using calculus.

First we note the solution x=1 by inspection.

Next, if 0<x<1, we have xˣ<1 (positive power of number in ]0, 1[), but 1-x²>0 so 2^(1-x²)>1, hence there is no solution for 0<x<1.

Finally, if x>1, we have xˣ>1, but 1-x²<0 so 2^(1-x²)<1, hence there is no solution for x>1.

So we conclude that for x>0, x=1 is the only solution.

For x=0, we have xˣ=1, but 2^(1-x²)=2, so x=0 is not a solution. This is still true if you prefer to regard 0⁰ as being undefined.

For x<0, I not going to give a rigorous treatment.

Now xˣ=2^(1-x²)
⇒|xˣ|=|2^(1-x²)|, though the converse is only true if xˣ and 2^(1-x²) have the same sign, i.e. if xˣ>0.
⇔|x|ˣ=2^(1-x²), since |xʸ|=|x|ʸ for real-valued real powers of real numbers, as long as xʸ is defined as a real number, and since 2ˣ is always >0.
This equation has the obvious solution x=-1, but x=-1 is not a solution to the original equation, since (-1)⁻¹=-1<0.


The graph doesn't show clearly what happens as x→-∞, but as |x|ˣ=2^(1-x²) has -x² in the exponent, it converges to 0 much more quickly than xˣ as x→-∞ (for example, for x<0 their natural logs are xln(-x) and (1-x²)ln2, and it can be shown applying de l'Hôpital's rule a couple of times that xln(-x)/[(1-x²)ln2]→0 as x→-∞), which makes it plausible that the two graphs never intersect for x<0. (I said this wasn't going to be rigorous!)

Hence we conclude there are no solutions to the original equation for x<0, and x=1 is the only real solution.

[anupchattaraj]

Very nice method.sir is it possible to solve the same by without help of calculus?

[giuseppemalaguti]

Ponendo x=e^u, risulta dopo vari calcoli.... shu/u=1/(ln1/4)<0....essendo le due funzioni shu e 1/u nel 1 e 3 quadrante il loro prodotto non può essere negativo, percio non ci sono soluzioni.... Rimane da verificare u=0.... Infatti u=0, cioe x=1 é l'unica soluzione