VietNam Math Olympiad

  Рет қаралды 2,335

OnlineMaths TV

OnlineMaths TV

Күн бұрын

Today's math problem is a VietNam Math Olympiad problem.
In this question, we are asked to find the sum of all the real roots to this Olympiad logarithmic equation.
Watch how Mr. Jakes skillfully dissected this VietNam math Olympiad problem with ease.
Watch other nice logarithm equations from this channel via the link;
/ @onlinemathstv
Kindly share this VietNam math Olympiad with VietNam College and University students using the link;
• VietNam Math Olympiad
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Пікірлер: 18
@ToddKunz
@ToddKunz 2 ай бұрын
I loved what you did here. Thanks for the great video.
@WorldwideBibleClass-qr9jk
@WorldwideBibleClass-qr9jk 2 ай бұрын
What a nice problem with a nice, unique and detailed/explicit explanation from Onlinemathstv. More wins sir....❤🎉
@charlesmitchell5841
@charlesmitchell5841 2 ай бұрын
Nice problem. Was a little bit of work. Well explained, thanks for the lesson and video. 👍
@SidneiMV
@SidneiMV 2 ай бұрын
log₃(5 - 3ˣ) + log₃3ˣ = 0 3ˣ(5 - 3ˣ) = 1 3ˣ = u => x = log₃u u(5 - u) = 1 u² - 5u + 1 = 0 u = (5 ± √21)/2 *x = log₃[(5 ± √21)/2]*
@legendarysquad8778
@legendarysquad8778 2 ай бұрын
nice! I was also thinking about this
@abdulkaiummondal9193
@abdulkaiummondal9193 Ай бұрын
@FERNANDOALVAREZ-iz8nw
@FERNANDOALVAREZ-iz8nw 2 ай бұрын
Excellent video, greetings from Monterrey, Mexico.
@antoniogomesfigueiredo7835
@antoniogomesfigueiredo7835 2 ай бұрын
Bonita questão, hein professor? Gostei do seu desenvolvimento. Ainda que eu não entenda bem o Inglês, porém, dá para acompanhar a sua explicação. OK? Deus abençoe você professor.
@viktarkrylov5604
@viktarkrylov5604 2 ай бұрын
t=3^x, 0
@amitsrivastava1934
@amitsrivastava1934 2 ай бұрын
The domain of the given expression clearly indicates that 3^x < 5. You cannot have ( 5 + √21)/2 as a value of 3^x. Am I missing out on something here ? Kindly review.
@zeroone7500
@zeroone7500 2 ай бұрын
(5 + √21)/2 < (5 + √25)/2 (5 + √21)/2 < (5 + 5)/2 (5 + √21)/2 < 5
@amitsrivastava1934
@amitsrivastava1934 2 ай бұрын
@@zeroone7500 Thank you so much. I had missed out on this.
@JC-rl5vy
@JC-rl5vy 2 ай бұрын
Nice chalk :), had fun with this
@onlineMathsTV
@onlineMathsTV 2 ай бұрын
Hahahahaha.....😍😂😂🤣😍🤣🤣 Thanks sir
@mallikarjunaraokotikalapud4828
@mallikarjunaraokotikalapud4828 2 ай бұрын
Yes nice problem.
@damyankorena
@damyankorena 2 ай бұрын
Alternative sol: Note: log will signify a base 3 logarithm for the rest of the problem. Let x=logy log(5-3^logy)+logy=0 log(5-y)+log(y)=0 log(5y-y²)=0 5y-y²=1 y²-5y+1=0 Σх=Σlogy=logΠy. The product of y1 and y2 by vieta is 1, therefore the sum of roots for x is log1=0
@onlineMathsTV
@onlineMathsTV 2 ай бұрын
Wonderful approach from you sir.
@prollysine
@prollysine 2 ай бұрын
/// if 3^x>5 , x=(ln(5+V21)/2))/ln3 or x=log₃((5+V21)/2) not a solu ///, solu , x=((ln(5-V21)/2)/ln3 , or x=log₃((5-V21)/2) ,
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