Video

and m=1?

At 2:58 It is mentioned that each factor is 0. I feel it should be either one of the two factors is 0

m = m/18; 18m = m; 17m = 0; m =0. Done. Typed that in 10 seconds.

BOOM CASE CLOSED

Please give me back 3 minutes of my time

Or you could just multiple it by 18 and distract right side from left

At 1:00 the solution is ready

m. =. 1/3

It can also be solved by using Lambert's W function.

big mistake at 2:25 a^a=b^b -> a=b is false. For exemple, if a=.739534 and b=0.09465 then a^a=b^b=.8 (approximation of course) But in your case, it's correct because 27>1 !

At the start there is no stipulation that n and m are both positive and Real. Without those constraints, x and/or y could be negative if m and/or n are Complex and in consequence, at 2:13, it cannot be asserted that x+y>x-y. Because of this, at 2:21 the assignment of 17 to x+y and 1 to x-y cannot be made without further justification. Swapping those assignments leads to the alternative of x=9, y=-8. At 4:40 one possible solution, m^n=3^4, is presented as a partial answer although other partial answers could be valid here (e.g. sqrt(3)^8) and at 5:20 m^n=4^3 is presented as another partial answer with the same caveat. It's simply convenient to state that m=3 and n=4 but there needs to be more work to prove that's the only solution.

i think there are 4 valid solutions to the system of equations, and two are shown in this video. the other two are... m = n = (1 + sqrt(965))/2 =~ 16.032 and m = n = (1 - sqrt(965))/2 =~ -15.032 if we graph the two quadratic equations in the problem on the euclidian plane, one is a parabola that opens upward and the other is a parabola that opens to the right, and there will be a total of four points of intersection

How you extend this question,, but your method is typical for students make it more easy Thanx from lala Pakistani

8:15 the ± sign is a bit confusing, you should use ∓ there to show that they're opposite signs

this one is easy sorry for the notation it is not pleasant to do math on e keybord (MN)^0,5=10 M^0,5 + N^0,5 = 10 (MN)^0,5=10 -> M=10^2/N (10^2/N)^0,5 + N^0,5 = 10 10/N^0,5 + N^0,5 = 10 (10 + N)/N^0,5 = 10 -> (10+ N)/10 = N^0,5 (10/10)^0,5 + N^2/100 = N 100 + N^2 = 100N -> N=(100+-(100^2-4 * 100)^0,5)/2 N= 98.9897948557 V N=1.01020514434 M=10^2/N M= 100/98.9897948557 V M=100/1.01020514434 M=1.01020514434 V M=98.9897948557 M=1.01020514434 and N= 98.9897948557 V N=1.01020514434 and M= 98.9897948557

2:53 why did (x-3) get common?

mn = 19.40512....

That's a fun one. I think it's important to state in the solution that m and n must have opposite signs on the second terms. I tend to write this by flipping the plus/minus sign on one of them Also you can sort of skip that last long bit of working - since m and n are symmetric in the original equation, the answer must also have symmetry: m can only take the values we've shown n takes

" Unlike 2^x = x, equation 2^x + x = x has one real solution: x = 0! #math #complexnumbers #mindblown" << one of the answers A.I. generates -.- the real answer is as transcendental as the variable i. shall we call it a non-existing number ?

(√m - 1)(√n - 1) = 10 - 10 + 1 = 1 Now (√n - 1) -> x and by sym. in the 1st eq: x + 1/x = 8 => √m = √n = 1 + 4 +/- √15 => n = m = 40 +/- 10√15

Let a = sqrt(m) and b = sqrt(n) a + b = 10 and ab = 10 So we are looking for the solutions of x^2 - 10x + 10 = 0 x = 5+-sqrt(15) x^2 = 40 +-10sqrt(15)

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Hmm, you have quite the handwriting, but I still think it would be better if you actually spoke. Good that you actually verified this, I don't see many people actually doing this.

a^a=b^b does not always imply a=b, e.g. (1/3)^(1/3)=x^x has also solution that is approx x=0.403542... But if a or b is at least 1 then this works, so we are good there.

No idea why this got recommended to me but I enjoyed it a lot, very satisfying

dude, does your camera not focus ?

Thank youuu🎉

As an old math olimpician, I think the question shouldnt belong to math olympiad exam. The problem was really easy.

5 fond in 5 secondes

Nice trick

Why not go to log in base 3 immediately on both sides at 3^m =6? You immediately get to m = log_3(6).

4^m=40 m=log base 2(40½)

Easy

why the solution is so long 2^(6 m)= 2^(2) * 2^((3/4) 2^(6 m)= 2^(11/4) 6m = 11/4 m= 11/24 it is really that easy i solved in my head under 1min

Let x=2^m. The equation becomes: x - 1/x = 4 x²-4x-1=0 x²-4x+2²=5 (x-2)²=5 x=2±sqrt(5) Since x>0, take only positive result. m=log2(2+sqrt(5))

Substitute x=2^m into the given equation: x-1/x-4=0. Multiply by x and rearrange to x^2-4x-1=0 with roots x=(4±2√5)/2=2±√5 Using the positive root, m=ln(2+√5)/ln2=2.0827...

Once you divide by 2i are you also left with m/2i ?

And x = ( by definition ) m

Nice but you missed a solution, also m= 8i

With PEMDAS multiplication and division are done by which comes first. So they can be performed out of order as far as the word. Meaning you would do the division ( 6 divided by 2) then times 3 { 3(3)=9. So the answer would still be 9 either way. M

I would think that either answer is correct depending on which order you were taught to use. I was taught PEMDAS so in my school of thought 1 is correct. If you were taught BODMAS then 9 is correct. I'll stick with what I was taught thanks but it's always interesting to consider other methods of thinking. I try to keep an open mind ....🥸

“1” - PEMDAS rules!

Bodmas

'Promo sm'

when constrains are b>=0 hit and trail mode on 😂😂😂😂

It's not the only one solution. a = 5 and b = 1 is also a solution. 5 + 2*5*1 + 1 = 5 + 10 + 1 = 16

This is only true if a and b are required to be integers. (You never specified that.)

Actually this is only ONE solution. There are infinite solutions since e^(i(pi + 2 k pi)) = -1 for all integers k.

10*9*8*7

Song ? Such a calming video