x=1 by observation

Bas... Rd....

Excellent!!!!

You could simply say: Let a²=x^y, b²=y^x. You'll arrive at the same answer, and it'll be faster.

128 is the key. Its 2 to the seventh power. Trying powers of 2,we find the answer in about a minute. That how a mathlete would solve this equation.

This is so beautiful. 🙏❤

Thank you sm ! This was easy that i thought i was wrong hehe May you solve more hard equation?

Me suscribo. 3 minutos sin calculadora. Bonito ejercicio.

Nice Explaination ❤.

The most important of problem solving is the plan. We look at the given equation and see 25 in two places. How can we use this information? We can rewrite it as.....And then you do the grunt work.

👌🏽

Ce système revient à résoudre x+10/x=10 En multipliant par x on obtient une équation quadratique qui a donc au plus deux solutions. Si on trouve deux solutions il n y en aura donc pas d autres. Ces solutions peuvent s ecrire sous la forme (a+sqrt(b))/2c et (a-sqrt(b))/2c Supposons c=1/2 Les solutions sont alors sous la forme a+sqrt(b) et a-sqrt(b) Avec : a+sqrt(b) + a-sqrt(b)=2a=10 soit a=5 Et (a+sqrt(b)) (a-sqrt(b))= (5+sqrt(b)) (5-sqrt(b)) = 25 -b = 10 soit b=15 Donc 5+sqrt(15) et 5-sqrt(15) sont solutions. Ce sont donc les seules d après ce qui a été dit plus haut.

Look at the equation!!Its clear one solution is 1/2.Since imaginary solutions occur in pairs,,1/2 is a double root. We can use synthetic division to find the other two roots.However,your solution is interesting and a learning situation.

Look at the equation. We have the sum of a number and its cube is 68.4 is obvious. No need for any explanation. The rest is trivial.

Nice! I got 4^(1/128)=x^(1/x^2) ie 16^256=x^(1/x^2) ie x=16

great problem

My method: 2x log 25 = log 50 2x = log 50/log 25 x = log 50 / 2 log 25 x = log 50 / log 625

Same value, i got answer x = log 50/log 625

Nice idea.

I wonder if this method is applicable eg to this : x^y - y^x = 399 ? or ...= 169

Taking the 4th root is not giving you the whole picture. You are looking at finding 4 solutions here. Video shows clearly that we have 2 real solutions and 2 complex solitons. a=1/2 is a repeated root based on the concept of discriminants.

We find it using squareroot as so- (a^2)^2=((a-1)^2)^2 square root of both sides a^2=(a-1)^2 Using (a-b)^2= a^2+b^2-2ab a^2=a^2+1-2a a^2-a^2+2a=1 2a=1 a=1/2

4^(x^2)=x^(128) 4^(x^2)=x^(2×64) 4^(x^2)=(x^2)^(64) Let y=x^2>0 4^y=y^64 y=4^(y/64) y/(4^(y/64))=1 ((-2yln(2))/64)/(4^(y/64))=(-2ln(2))/64=(-8ln(2))/256 (-2yln(2))/64=-8ln(2) y/32=8 y=256 x²=256 x=±16 OR (-2yln(2))/64=W_0(-ln(2)/32) (-yln(2))/32=W_0(-ln(2)/32) y=(-32W_0(-ln(2)/32))/ln(2) x²=(-32W_0(-ln(2)/32))/ln(2) x=±sqrt((-32W_0(-ln(2)/32))/ln(2))

4^(x²) = x¹²⁸ = (x²)⁶⁴ x² = u 4^u = u⁶⁴ uln4 = 64lnu (1/u)lnu = (1/64)ln4 (1/u)ln(1/u) = (1/64)ln(1/4) (1/64)(1/k) = (1/4)^k 64k = 4^k 4^(k - 3) = k => k = 4 (by inspection) (1/u)ln(1/u) = (1/64)(1/4)ln(1/4)⁴ (1/u)ln(1/u) = (1/4⁴)ln(1/4⁴) u = x² = 4⁴ => *x = ± 16*

It's even worse than just trial and error, there are infinite solutions Anything where 9^x>17 will be part of a valid solution.

The best methos ever. Guess & check is not a real method. Thank you for showing us this :)

Can't we just underoot 4 on both the sides??

Nice !!!!

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Don't work so hard! 18^1 - 1^18 = 17

3^4-4^3 easy in brute force mode like y made at the end. BUT is not the only one...

Why going through all these elevations/simplification? The lhs writes: (2^2*(2*2^1/2)^1/2)^1^x. Starting from the inside and expanding, step-by-step, toward the outside parenthesis using basic power formulas (to simplify notations, I wrote a^b/c instead of a^(b/c) as it should have been), you get: 2 * 2^1/2 = 2^(1+1/2) = 2^3/2 (2^3/2)^1/2 = 2^(3/2*1/2) = 2^3/4 2^2 * 2^3/4 = 2^(2+3/4) = 2^(8/4+3/4) = 2^11/4 (2^11/4)^1/x = 2^11/4x (where 11/4x is to be read as 11/(4x) ) In other words, the original equation can be rewritten as 2^11/4x = 2^5. As exponentiation is injective, this is equivalent to 11/4x = 5 and, thus, 11 = 20 x, i.e. x=11/20. This approach is of course equivalent to yours but, in my opinion (I'm open to contradiction'^^) is "simpler" in the sense that it only involves applying formulas minimizing the reasoning of the form "elevate both parts to ..." that (seems to me) tend to be confusing to "beginners" P.S. The "As exponentiation is injective" argument is there only for the sake of being rigorous in my post's argumentation and should be ignored by whoever doesn't know what an "injection" is and considers that thee equivalence 2^11/4x = 2^5 <=> 11/4x = 5 obvious ;)

visit "Explored Maths" for very interesting problems @explored maths

Don't forget there's infinite solutions as u can endlessly add 2pi or 360 if degrees

Isn’t this a function？

We have a cubic equation: x^3-x^2-36=0. It has an obvious solution x=-3. By factoring, the cubic equation becomes (x+3)(x^2-4x+12)=0. The second factor gives two additional komplex solutions x=2(1+i sqrt(2)) and x=2(1-i sqrt(2)). There is no need for lengthy mantal gymnastics...

x= 11/20

решил в уме 8*8 = 64, 64 -39 = 25

The other solutions are x=1, y=-16; y=1, x=18.

There is an easier way to get this answer. 25^(2x) = 50 25^(2x) = 2 * 25 ( 25^(2x) ) / 25 = 2 25^(2x - 1) = 2 log₂₅(2) = 2x - 1 x = ( log₂₅(2) + 1) / 2 It's the same thing, but much shorter :)

60 degrees and 30 degrees is correct

"It is proved that the values of x and y are correct". Yes, after a very very long and painfully tedious series of steps you ha arrived at a solution. But the problem has many more solutions, in fact it has an infinite number of solutions. The trivial solution that you have found is obvious after a quick glance at the problem: 39 is 64 - 25, 64 is 8^2, 25 is 5^2, therefore x=16, y=16 is a solution.

Man, I knew how to do that, but I am pretty sure I am gonna fail it if I meet it anywhere, huh. Respect to you, brother, for making videos like this

x=11/20

Fantastic

Why not (a+b)(a-b) = 39.1 ?

X^y-y^x= 17 3^4-4^3=17 81-64=17 X=3 Y=4

X^1/2=8^2 X=8^4= Y=5^4

Wish you good luck.....keep uploading this type of content 🎉

in degrees: 60 + 180n, 120+180n, 30+180n, 150+180n