Math Olympiad Question | Equation solving | You should learn this trick to pass the exam

Equation solving with best tricks ever and excell in exams!
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Maths Olympiad is an examination curated with advanced level mathematics to work on a student's current potential. In such tests, the students are required to solve the given questions. Based on the problem solving, the student is rewarded with marks and accolades. To support this we provide with solutions for the toughest and all level of questions.
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Benefits of Math Olympiad
* Fantastic exposure to the level of complicated mathematics.
* Great opportunity to test you whether one is ready for the said complicated mathematics.
* If someone is passionate about mathematics, Olympiads will undoubtedly be an extraordinary chance to stay in touch with your passion.
* Olympiads can also be an excellent way to prepare for the future if someone is looking ahead to take up mathematics for their higher studies.
* Olympiads give a chance to get acquainted with your weaknesses and strengths and work on them.
* Since Olympiads happen quite a few times, around the year, this can help keep those skills under practice.

[@phoebe543]

For everyone who thinks she's wrong here's an explanation. The issue here is that when she introduces the rule in the first step in her solution (a^m)^n = a^m×n but the problem on the page is a^m^n = a^(m^n) She also says "whenever" you have a power to a power you multiply the powers (which is not true) so it appears that she's going to treat the problem as x^4x but she's actually introducing the rule because she's about to raise (x^x^4)^4 and then rewrite it as (x^4)^x^4 = 8^8 So, x^4 = 8 x = 4root8 She did a very bad job of explaining and seems to have mispoken so it's got people more confused than before watching the video. So possibly not the best teaching moment!

[@jim2376]

The key is in raising to exponent 4 on both sides, the 4 on the LHS goes on the x downstairs for (x^4)^(x^4). So we've got (x^4)^(x^4) = 8^8. Then it's application of a^a = b^b having equal bases. x = 8^(1/4).

[@VedhGajawada-jt3ri]

I think this is a good way for understanding the problem because some people think she's wrong which is true in some parts, as she was incorrect, but she ended up getting the right answer.

[@arungupta3509]

There is an ambiguity in the problem. The solution given is for x raised to (x raised to 4). But what about (x raised to x) raised to 4? Note that the brackets matter. E.g., (5 raised to 3) raised to 2 is (5 raised to 6), while 5 raised to (3 raised to 2) is (5 raised to 9).

[@faisalshaikh8741]

exact I came here to say that bracket matters.

[@ciaranmcnulty5713]

@@faisalshaikh8741 if there are no brackets it is defined as x^(x^4) - you only must add the brackets if you want (x^x)^4

[@donaldwebster1533]

​@@ciaranmcnulty5713Even if there is such a convention there is still an unacceptable ambiguity - witness that the problem is presented without brackets and solved as (x^x)^4 (with brackets by your convention).

[@Lalalalalalalall4093]

That's what I was thinking

[@mikebrau5354]

The standard notation is to compute the deepest exponent first.

[@sharonnoel2218]

Take log to the base 64 and differenciate wrt x

[@ftLabib-nc5pw]

Tthe math is totally wrong

[@Fernandez218]

she is correct. x^x^4 . plug in fourth root of 8, or 8^(1/4) . starting from the top you start with [8^(1/4)]^[8^(1/4)]^4. next you get [8^(1/4)]^8. then you get 8^2. finally it's 64.

[@romandboiko]

@@Fernandez218 the problem is that the author interpreted the left side of the equation as if it was (x^x)^4. But the notation actually means x^(x^4) and this isn't equal to x^(4*x).

[@Fernandez218]

@@romandboikosomehow she got the right answer.

[@mikebrau5354]

No, she did it correctly. x^(x^4)

[@romandboiko]

@@mikebrau5354 yes, you're right. I stopped watching at 0:18, thinking that the next step would be claiming that the left side is x ^ (4 * x), which would obviously be incorrect. After watching the whole video, I agree, she did it right.

[@mrmimi807]

You write the equation wrongly It's supposed to be (x^x)⁴ not x^x^4 If we assumed x = 3 then the first one equals: (2^2)⁴ = 4⁴ = 256 And the second one equals: 2^2^4 = 2^16 = 65256

[@binae8581]

It easy with lambert function

[@emreyurtoglu3283]

Math left the group.

[@abcd91208]

This is wrong....the question is x power x power 4 which is different from x power x4

[@DavidChristman-q7l]

x=4th root of e^W(4ln64)

[@DavidChristman-q7l]

using wolfram alpha that equals 2^3/4 (2^3/4)^(2^3/4)^4 does indeed equal 64 lambert w op🗿

[@mathiq56]

How i can improve my channel early?

[@BenJehovah6969]

X=EAZY-E and NWA

[@BaldeMamadou-o9n]

X^x^4=64 X^x^4=2^8 X^x^4=2^2.4 X^x^4=(2^2)^4 equi à x=2 Véri: x^x^4=2^2^4😮😮😮😮😮😮😮😮😮😮

[@aniksamiurrahman6365]

Nice🎉

[@LKLogic]

Thanks 😊

[@justplayit2533]

1:09 This step is completely wrong

[@mech_eng_tech4468]

your method is wrong

[@MalluTech123]

Answer is 2

[@alvarosteiger]

And -2

[@superman00001]

Nooooo….. 4th root of 8 is not +/-2. Think again!

[@alvarosteiger]

@@superman00001 Yeeees .... x = +/- (4th root of 8) so x = +/- 2. See the video again.

[@superman00001]

@@alvarosteiger 🤣 the fourth root of 8 is not +/-2

[@alvarosteiger]

That's why you are @Superman 😆

[@TruiluiTran]

Very not good

[@vikramsharma5618]

Wrong. The powers are at different level and you cannot multiply them. Your logic is correct but your notation is wrong.

[@mikebrau5354]

The notation is correct. x^x^4 = x^(x^4)

[@romandboiko]

Consider revisiting the entire video; its progression diverges from the initial impression. The presenter engages in preparatory steps before multiplying the powers.

[@ShedrachCodded]

I know it is correct, but I'm confused totally of the way it is solved

[@rickymort135]

The annoying thing is she just says "now I'm going to this..." without explaining the rationale of why you would think to do that. This means while you can follow it you can never apply it which should be an important thing to reach. Unfortunately too many math teachers do this. This here seems like a pure observational trick where raising everything to the power of 4 results in the right being 8^8 and left being []^[]. Change the 4 to a 3 and it doesn't work...

[@romandboiko]

@@rickymort135 numerous olympiad questions share this characteristic - it's crucial to notice a key aspect of the problem to uncover a solution. In this case, the simplification involves a strategic alteration of the unknown. Given that the first step to be calculated is x^4, it makes sense to introduce that as a new variable. The challenge lies in eliminating the original x. Employing a fourth power operation on both sides proved instrumental in achieving this.

[@pauldalnoky6055]

Me too. Way too fast😊

[@superman00001]

Well done, lovely woman.

[@LKLogic]

🥰

[@rangde_colours_of_fashion]

Simp

[@salamatuibrahim3919]

I am agree with you

[@steffenhantschel2016]

The trick is at 1:30, but you don´t get any hint before, so you couldn´t understand the steps before.

[@MADARAGAMING_FF-GU4]

Wrong

[@haiyangwan2363]

Whilst the written narrative shows what the speaker is thinking, some of the spoken narration is incorrect (eg "64 can be written as 8 squared OVER 4") which makes it a bit difficult to follow . In addition at 1:38 saying "now we can compare either the bases or the powers" and then going on immediately to say "Thus we have X^4 = 8" is moving a bit fast. Nothing is said about what was found as a result of making this comparison (ie that both the bases and the exponents are the same on both sides of the equation). It seems to be taken a obvious that if a^a = b^b then a = b. Whilst a=b is one solution to this, it is not obvious (to me at least) that there are not other solutions. Does anyone have a proof that a^a = b^b implies a = b?

[@GilanClips]

a^a = b^b does not imply a = b. You can see a=1/2 and b=1/4 satisfies a^a = b^b.

[@haiyangwan2363]

@@GilanClips Thanks. So it appears the speaker has not proved that the solutions they give are the only solutions to the equation, as they have not proved that x^4 = 8 is the only solution to (x^4)^(x^4) = 8^8.

[@cosmoslover9855]

Can we do it by taking log

[@sqrtminusone]

yes you can but you need to use lambert W function, but then we are in complex numbers and i think she meant real numbers

[@DavidChristman-q7l]

yea x^x^4=64 x^4lnx=ln64 let y=lnx then x=e^y (e^y)^4*y=ln64 y*e^4y=ln64 now we almost have the form W(x*e^x)=x multiply both sides of the equation by 4 to get 4y*e^4y=4ln64 now we can use W🗿 W(4y*e^4y)=W(4ln64) 4y=W(4ln64) remeber we let y=lnx so we have 4lnx=W(4ln64) rewrite using power rule for logs lnx^4=W(4ln64) e^lnx^4=e^W(4ln64) x^4=e^W(4ln64) (x^4)^1/4=(e^W(4ln64))^1/4 x=(e^W(4ln64))^1/4 or x=4th root of e^W(4ln64) using wolfram alpha to check this value it gives 2^3/4 (2^3/4)^(2^3/4)^4 this indeed does equal 64🔥

[@k.lakshmi1406]

😂😂😂 Simple math Consider x power 4 = a x^a = 2^8 Then a = 8 So , finally x power 4 = 8.

[@user-대구선비]

In real number, it is correct. But In Complex number, we need to have 2 addtional solution. x = +-(8)^^(1/4), +-(8)^^(1/4)i

[@7thclasstopper952]

X=1/4

[@a.bofficial8945]

Answer is 2

[@sevaksirmathofficial8961]

Madam you are false

[@JaredObiero-b1m]

this is very true

[@sevaksirmathofficial8961]

Ap ne galat batya hai

[@thebeast9488]

Plz show your face

[@vstechiegamer5996lol]

Why

[@thebeast9488]

​@@vstechiegamer5996lolwho are you

[@vstechiegamer5996lol]

@@thebeast9488A person?

[@cooldudehfgd584]

Creep

[@thebeast9488]

@@cooldudehfgd584 are you talking about yourself