VietNam Math Olympiad
Вставка
- Опубліковано 9 вер 2024
- 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
#vietnam #school #maths #trending #education #challengingmathproblems #olympiad #olympiades #challenging #log #logarithm #solution #viralvideo #viralmaths #viralshorts #viralshort #viralvideos #viralreels #viraltiktok #viral_video #everyone #everyday #everydaymotivation #evergreenhits #evergreen #always #guru #mathematics #math #mathstricks #mathura #mathtrick #mathshorts #solution #solutions #always #alwaystrends #how #howto #howtomake
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]*
nice! I was also thinking about this
❤
t=3^x, 0
Excellent video, greetings from Monterrey, Mexico.
What a nice problem with a nice, unique and detailed/explicit explanation from Onlinemathstv.
More wins sir....❤🎉
Nice problem. Was a little bit of work. Well explained, thanks for the lesson and video. 👍
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
Wonderful approach from you sir.
I loved what you did here. Thanks for the great video.
Yes nice problem.
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.
Nice chalk :), had fun with this
Hahahahaha.....😍😂😂🤣😍🤣🤣
Thanks sir
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.
(5 + √21)/2 < (5 + √25)/2
(5 + √21)/2 < (5 + 5)/2
(5 + √21)/2 < 5
@@zeroone7500 Thank you so much. I had missed out on this.
/// 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) ,