Math Olympiad Question | Solve the Radical and Exponential Equation | Math Olympiad Training

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Math Olympiad Question | Solve the Radical and Exponential Equation | Math Olympiad Training

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[randomjudgements]

Very good sir that was a great video
Thanks for the great solution proffesor 😀😀
And you are ao close to 300k 👍

[christianthomas]

just factorizing under the radical to obtain after simlification: V((2^x) - (3^x)) = V(3^x) then, after squaring the two members you get after some obvious simplification: 2^(x-1) = 3^x and aafter logging the two members you get: x= (log(2))/(log(2/3))

[bigm]

Very smooth transition from bewilderment to enlightenment.

[marcod]

Sir, it’s always interesting to me to see different approaches in solving the same problem! Thank you! My way would be this:

6^x - 9^x = (3^x)(2^x - 3^x), and it must be at least 0 (since we want its square root), which happens whenever x<=0. So: let’s set
2^x - 3^x = y^2, with y>=0 for simplicity and no loss of solutions.

Thanks to the substitution above, we have:
y^2 = sqrt(3^x)*sqrt(y^2)

So:
y(y-sqrt(3^x))=0

Either is:
y=0 ->
y^2=0, and, remembering how we defined y^2, ->
2^x - 3^x =0 ->
x=0 (first solution, non positive and then acceptable), or is:
y-sqrt(3^x)=0 -> y=sqrt(3^x) ->
y^2=3^x ->
2^x - 3^x = 3^x ->
2^x = 2*3^x ->
2^(x-1) = 3^x
and then, taking the logs:
(x-1)*log2=x*log3 ->
x*log2 - log2 = x*log3 ->
x*(log2 - log3) = log2 ->
x = log2 / log(2/3) ->
x = log[2/3]2, with which notation I mean log of 2, base 2/3.
This is the other solution, non positive and then acceptable.

Just 4 fun. Bye.

Edit: I wrote the solutions are non negative, whilst they’re obviously NON POSITIVE and then acceptable. So sorry for that. Beg you pardon.

[ACheateryearsago]

This was a hard one, couldn't have solved it without your help..

[murdock]

This was a great one, many thanks! :-) Learned a lot! You are Master of Math!

[HappyFamilyOnline]

Very nice step by step tutorial👍
Thanks for sharing🎉

[mathsplus]

Thank you for your video solution for radicals

[Copernicusfreud]

I ended up with the same answer. I solved for x ^2 first, and then multiplied by x to get x^3, which is -1. x ^ 7777 = x * (x^3)^2592, so x^7777 = x.

[owlsmath]

Nice one! I did it basically the same way except i used log base 2 although that makes it a bit more messy. Like your way better! 😁

[hansschotterradler]

Fascinating. X=0 is the obvious solution.

[hiwaforlife]

Thank you sir, it was informative, keep it up.

[seegeeaye]

Let y=2^x – 3^x, then the given equation becomes y^2 = 3^x (y), from which we have
y=0 or y=3^x, then we get the two solutions as in the video

[derhausfreund]

c := sqrt(2^x - 3^x) usw. --> quickly

[hanasuajipi]

podstata reality je rozumné uvažování v dané situaci jak se člověk zachová zmáčkne spoušť, nebo se to pokusí zachránit a zabrání i vlastním tělem co se dá .

[GaetanoCoiro]

la differenza tra 6^x e 9^x non é sempre positiva da poter giustificare la semplificazione tra radice quadrata e quadrato del secondo membro.
6^x é maggiore o uguale di 9^x per x<=0.
Bisogna aggiungere perció la condizione : x<=0.
Diversamente, la semplificazione indicata non é corretta.

[xyz]

Not sure the substitutions and squaring are helping or made the problem look more complicated than it was.

[praneelramjutton]

X=O, is not a real solution for this equation

[Raj_IIT_KGP]

I solve it in 1.5 min worst explanation 🤣🤣🤣..this method is big..

[TheLukeLsd]

I found log⅔ (2) or log 2÷log ⅔. And the case of 0 considering the a-b =0 where 2^x=3^x is just possivel for x=0.