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First impression: x = 2 is an obvious solution. If we sketch the graphs y = 5^x and y = 50/x it would seem there can only be the one intersection, therefore x = 2 is the only solution. Now I'll watch the video. I am sure there will be some subtlety that has escaped me. This is a wonderful channel.

X can be (5)^(1/4), -(5)^(1/4), (5)^(1/4)i, or -(5)^(1/4)i

At 2:12 I feel like you should make sure that y^2-y is not 0 before cross multiply. It can happen only if y = 0 or y =1. y cannot be 0 and y = 1 can happen if x = 0. Luckily you did not find x as a solution but we must always be careful!

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let u=2^x , -> u^3-5u^2+6u=0 , u(u^2-5u+6)=0 , /// u=0 , 2^x=0 , not a solu. /// , u^2-5u+6=0 , u= 3 , 2 , u=3 -> 2^x=3 , x=ln3/ln2 , u=2 -> 2^x=2 , x=ln2/ln2 , x=1 , test , x=ln3/ln2 -> (8^(ln3/ln2)+4^(ln3/ln2))/(4^(ln3/ln2)-2^(ln3/ln2))=36/6 , 36/6=6 , x=1 -> (8+4)/(4-2)=12/2 , 12/2=6 , OK ,

x = 2

I got x=2 right away, but I could sense Lambert's W function being used just by looking at the original equation.

Two is clearly one solution. About two seconds to notice that.

"t"(tea) or coffee. Ha ha I love it! Better than to "b" or not to be!

"t"(tea) or coffee. Ha ha I love it! Better than to "b" or not to be!

x5^x = (2)5² xe^(xln5) = (2)5² (xln5)e^(xln5) = (2ln5)5² (xln5)e^(xln5) = (ln5²)5² (xln5)e^(xln5) = (ln5²)(e^ln5²) xln5 = ln5² = 2ln5 *x = 2*

Seems like this is overcomplicating it to me..

The function is strictly increasing and continuous, so it's invertible, so the obvious real solution is unique in the reals. But if you want complex solutions also, then lambert W is obviously the correct approach

x=W(50ln5)/ln5

Before even viewing, I smell the obnoxious odor of Lambert's W function.

After we guess x=2, we have to show that this solution is unique. Consider the function f(x)=x5^x. Note that for non positive x, f(x) is also non positive. and for positive x, f(x) is strictly increasing (no need to differentiate to see that). so there is a unique solution to our equation.

Why use Lambert’s W Function when you can easily express that 50 = 2 * 25 = 2 * 5^2, meaning x = 2. It’s a similar process relative to using the W(), but much faster 😑

The prime factorization of 50 is 2*5^2. By 1:1 correspondence x = 2. It’s not “guess and check”.

The exponents of the prime factors of a perfect square are even !

6 ! = 2 * (3!) * (3! ) * 4 * (5 !) ^2 * 6 = ( 4 * (3)! * (5!)) ^2 * 3 Hereby a * (6!) = b^2 implies a * 3 * ( 4 * (3!) * (5!)) ^2 = b^2 Hereby b = 4 * (3!) * (5!) * 3 = (2!) * (3!) * (6!) a = 3

impressive , i thought of this problem when i was first introduced to functional equation but i couldn't solve it.

😊😊😊👍👍👍🎉🎉🎉

i did this in my head in 1 minute as i looked at the thumbnail

6!=1*2*3*4*5*6=6*4*5*6=2*2*6*6*5=12*12*5 So if a*6!=b^2 with natural numbers a and b, a must be dividable by 5, because the lleft side of the equation is diviidable by 5, so b must be diviidable by 5 and b^2 must be dividable by 5^2. So a=5 and b=60 is a solution, also a=5 and b=--60. The others solutions are a=5*c^2 with a whole number c and b=60*c. There is no solution with a negative whole number,because the right side of the equationis can never be smmaler than zero.

It’s not a number theory problem; it is simple prime factoring.

Actually any a of the form 5^(2m+1)*c² will work.

5a x 12^2 = b^2 a = 5^(2n-1), b=12x5^n, non negative int n

you are very intense

a = 5, b = 60

Syber vs Geometry 1-0 🤩

Boo to the people that demand you edit out all errors. Math is often an iterative process mentally where you back up and try again and not showing that really isn't teaching math.

Very nice!

a=b=6! Surely?

盲猜 x^2-x+1=0 x^3=1

Why do you ramble on

Well, actually that is not quite true. a must be a multiple of 5, agreed. But it might be a multiple of any prime number greater than 5 raised to the power of 2n, where n is an integer. Every prime number might appear or not as a multiple and n can be different for every prime number...

8:14 I can hear Any Math saying "Nice".

Using CBS we get (x+y+z)(1/x+1/y+1/z)>=(1+1+1)^2=9 =)))

At first glance: 6! is 2^4 * 3^2 * 5^1. All perfect squares have all prime factors to even exponents. So any number of the form 2^(2k ) * 3^(2m) * 5^(2n-1) * p^2 (where k, m, n and p are non-negative integers) should accomplish this. (Edit: actually, we can simply say 5^(2n-1) * p^2, because the 2 and 3 can be rolled into the p.)

It is much simpler .. every digit in the faculty must be compensated by something to give a square. The 6 by 2*3 to give 6*6, the 4=2x2 is a square already so stripe away 2,3,4 and 6 (2*3*4*6 is square) on the faculty site and they need not be compensated anymore , the 5 cannot be compensated, so it must have factor 5 to give a square 5x5. So a = 5 and b=sqrt(4) *5*6 =60 . I did it without paper ,by head.

If a = (1/20) the b would be ±6 OR if a=(1/5) b= ±12, so I think there will be infinitely many solutions.

Bravo. It's not difficult, but it's interesting how you parametrize the solutions with the integers

yes!!!!! 👌

why leave that bit in?

Nice

a•6! = b^2 a•(3^2 • 4^2 • 5) = b^2 Let a= 5(x^2) (3•4•5•x)^2 = b^2 (60x)^2 = b^2 So b=60x, a=5(x^2)

1st method ok, 3rd method standard. 2nd method: Witchcraft

6!=720=5*144...a=5,a=20,a=45...a=80...

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There is not much difference between the 2 methods