วีดีโอ

And -9 too. When you take sqr a becomes a module.

How is this supposed to be an exercise of the Cambridge University? LOL

Consecutive squares always differ by an odd number, so if you set that difference to 49 you get the consecutive integers 24 and 25, but clearly this ordered pair solution is not unique

Fun problem! I used the properties of natural logs to get k = ln(50)/ln(25). However, your method is more easily verifiable by hand, which I like a lot for competitions.

number system? integers? ;) if you want oxford, be correct!

It's no fun to use a picture of Hilter to advertise this video.

I solved it by inspection but unfortunately I can't show workings. 😏

Please change this thumbnail - it's very very disturbing, misleading and not appropriate.

there is another answer x=0.127869 if you are good find that answer

Don't use a hitler picture promoting a "german olympiad problem" video. Would you add a picture of Hussain Muhammad Ershadf for a bangladesh olympiad problem?

a^a=b^b means a= b is false, i.e., a=1/2, b=1/4. Needs more care at that step (true if a>1)!

The rising function is equal to a constant, thus only one soulution exists. Then you can literally solve it by common sense and just putting in 3.

it's always a good sanity check to get a sense of the size of the solution before starting. The exponential function of real numbers is never negative so we're looking for solutions for x≥0. Consider the function y = e^(x^2-1) - x. When x=0, the LHS is 1/e ≈ 0.4, which is positive. When x=1, y = 0. That's a solution, but there may be another. When x=2, y = e^3 - 2 (about 18). When x=0.5, y = e^(-3/4) - 0.5 ≈ -0.028. That shows there is another solution between x=0 and x=0.5. Check x=0.4508 gives e^(x^2-1) ≈ 0.4508. I think you'll find Lambert-W is multivalued. (Aside: Note that the thumbnail asks for the solution to e^(x^2 - 1) = x^2, and that has the solutions x = ± 1.)

a=3. b=1

This doesn't constitute a proof that these are the only solutions in Z+. In this case, they are the only solutions, but in general with f(b)*g(a)=12 it is not true that only six cases need to be examined. It would be very easy to construct functions f and g such that 'a' and 'b' are in Z+ but f(b) and g(a) are not in Z+ and f(b)*g(a)=12. Given this is not a proof that these are the only solutions, I wonder why such lengths were gone to in the video, when it would have been straightforward to immediately read the solutions off the original equation...

Can't you just say: 5=4+1 4 can be written as 4/1 1 can be written as 3/3 4/1+3/3=5 4+1=5 a=3, b=1

Click bait for me man.

By inspection, x = 3. (Time taken = 2 seconds)

You went through a lot of work for nothing. 11 = 2ˣ + x = 8 + 3 = 2³ + 3 = 2ˣ + x ⇒ x = 3 You were extremely lucky that 2³+3 = 11. If it was 2ˣ + x = 13, it might not have been solvable by hand.

I'm guessing most people watching this can do this in their head and are watching to see if they've missed something.

We get the same answer just solving for k^4 = 16, why the extra steps?

!!!

Damn...I could've been a Stanford student

k=256 I solved it yesterday on a different channel, it was in fact quite easy - math problems which boil to base 2 are usually easy foe mev😂

"Tricky" or pointlessly tedious? I can't decide which!

Be f: R ---> R, x ---> 2^x + x = exp(x.ln(2)) + x. For any real x we have f'(x) = (2^x).ln(2) + 1 > 0, so f strictly increases on R. Consequence: if the equation f(x) = 11 has a solution, then this solution is unique. As x = 3 is evident solution, then it is the only solution of the given equation.

256

{(X^4)/4}+1={(x^4)+1}/4 how is it possible?. It may be{ (x^4)+4}4

(7/m)+(8/n) ‎ =  9. n,m are real positive numbers. This 2 variable equation has infinite solutions. You can substitute directly him the above equation, but for fun: (8/n) = 9- (7/m) (1/n) = (9- (7/m))/8 n = 8/ (9- (7/m)), ………………… (1) on the condition 9-(7/m) > 0 or m> 7/9 Substitute in (1); m = 1 n=8/2=4 m=7 n = 8/ (9-1) =1 m = 100000000000000 n less than 8/9

wow 56/9 wow

It is easier to look at each term in the sum for positive integer values of m & n. If m=1 then n=4 (maximum value and happens to be a valid integer solution in itself). Looking at the 2nd term, if n=1 then m=7 (maximum value that also happens to be a valid integer solution in itself). So we can conclude 1<=n<=4. We can quickly eliminate n=2 or n=3 by brute force as they don't produce integer solutions for m. So the only two possible integer solutions for (m,n) are (1.4) and (7,1).

It took me a second to figure out that N=4 and M=1 7+2=9

BRUH, i thought they asked find x and y, for a second i thought i was smart :(

2x2x2+2=10

Stupid problem. How is that longer expression any more enlightening than the original?

All you need to know this equation is that 8+2=10 :)

40^40+40/80^40=40^40×40^40/80^40 =(40/80)^40×40^40 (1/2)^40×40^40=(1/2×40)^40=(40/2)^40=(20)^40

SAT does not has Calculus 💀

Complete sledgehammer to solve a very simple problem. Square both sides x^2 = 3x.Sqrt(4x) leaves x = 6 sqrt(x). square again, x^2 = 36x. x=36.

I found another solution 40^80/80^40 = (40^2)^40/80^40 = 1600^40/80^40 = (1600/80)^40 = 20^40 :)

P = 6 . 👍👍👍👍👍👍👍

The music is OBNOXIOUS, made me run away

Don't forget the 2pi*n.

2:40 I thought it was pi factorial and got confused where did that come from. But turned out it was just pi ^ 1 * ...

The music very awesome

Or more simply -i

a^3 + a^2 = 36 a^3 + a^2 - 36 = 0 (a - 3)(a^2 + 4a + 12) = 0 a = 3, -2 +/- 2i√2

3

Bro why did you substitute v for 2 there is not need to do that because you ultimately made it into 2 in the middle anyways.

Is there a rule that says I can't take the cube root of each side of the equation first? And, wouldn't that limit it to a degree 2 (or perhaps 3) problem? Regardless, convention seems to dictate that you only search for real solutions unless the problem either demands imaginary solutions or the problem specifically asks for all types of solutions, right?