Mathematical Olympiad | Learn how to solve Olympiad Question quickly | Math Olympiad Training

[@FaixaPretanoENEM]

Always a pleasure watching your videos, professor! You're making difference, keep doing it, thank you. Hugs from Brazil 🇧🇷

[@PreMath]

You are very welcome. So nice of you. Thank you for your feedback! Cheers! You are awesome, CRMF. Keep it up 😀 Love and prayers from the USA!

[@TheLukeLsd]

Salve parça! Esse canal é maravilhoso, mesmo!

[@johngreen3543]

It seems a little long using base 10 logs. I did it in just a few steps 11^y =6 and 11 = 9^x substituting for 11 in the first expression gives you (9^x)^y = 6 so 9^(xy)=6 then 9=3^2 gives you (3^2)^(xy) =6 gives you 3^(2xy)=6 = 3*2 so divide both sides by 3 and you get: 3^(2xy)/3 =2 or 3^(2xy-1) =2 so log base 3 of 2 is the exponent on 3 that gives you 2 which is 2xy - 1.

[@hans-rudigerdrzimmermann]

Exactly how I did it in approx 40 seconds or less. Congrats!

[@hans-rudigerdrzimmermann]

One only gets a heart by this channel if one does not comment or critize the video solution. Not allowed!

[@pedroparra587]

Yo lo hice de otra forma. (Paso1) Llamemos la incógnita K = log3(2) se puede escribir como: 3^K=2 (i). (Paso2) Sustituimos (i) en 11^y=3*2 => 11^y=3*3^K => 11^y=3^(K+1) (ii). (Paso3) Sustituimos 9^x=11 en (ii) => 9^x^y=3^(K+1) => 3^2xy=3^(K+1) => 2xy=K+1 => K=2xy-1

[@xyz9250]

Did the same way, read your comment after positing mine. At least start with log_3 would be faster than converting log_3 2 to base 10

[@Mete_Han1856]

Here is a fascinating question and a patient solution. Thanks.

[@PreMath]

You are very welcome. So nice of you. Thank you for your feedback! Cheers! You are awesome, Metehan 😀

[@dicksonphisthur3398]

It's a privilege watching you work.

[@PreMath]

So nice of you. Thank you for your feedback! Cheers! You are awesome, Dickson 😀 Love and prayers from the USA!

[@anatoliy3323]

Nice a task and its solution step-by-step as well! Thank you very much, sir! I consider you as my the best math teacher! God bless you!

[@PreMath]

Dear Anatoliy, you are so generous! Thank you for your nice feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@KAvi_YA666]

Thanks for video. Good luck sir!!!!!

[@PreMath]

You are very welcome. So nice of you. Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@saidrabie5081]

thank you sir for your easy approach and steps..all understood..

[@ManjulaMathew-wb3zn]

Hi. Premath! I always enjoy your videos. You stay very calm a professional throughout the video always. On this solution it caught my eye right away the number 11 can be left out easily in the process. Substituting 9^y in place of 11 you get 3^{2xy)= 6. Taking log on both sides at base 3 you get xy = log Base3 (2) + 1. Moving +1 to the other side you get the answer as xy -1 I love how your using more geometry than trigonometry on many solutions that make it more visual than mathematical.

[@fouadhammout651]

Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal......

[@farshadfattahi20]

Great. Many thanks

[@HappyFamilyOnline]

Always amazing👍 Thanks for sharing😊😊

[@nicogehren6566]

very nice approach

[@SuperYoonHo]

love your channel prof.

[@michelg3000]

Thank you very, very, much. I honestly don't how to thank you, Professor. May God continue to look upon your life...

[@242math]

this solution was very well explained, thanks for sharing this Olympiad question

[@PreMath]

Glad you liked it You are very welcome. Thank you for your feedback! Cheers! You are awesome 😀

[@v-gc7257]

Very Well explained . Helpful indeed. Easy tutorial. Thank you for the effort of doing this

[@johnbrennan3372]

Very interesting method Than you.

[@nirupamasingh2948]

Vvvv nice explaining each step

[@PreMath]

So nice of you. Thank you for your feedback! Cheers! You are awesome, Niru 😀

[@tonyennis1787]

Very nice indeed

[@randiwijaya9609]

Interesting problem 👍

[@PreMath]

It sure is! Thank you for your feedback! Cheers! You are awesome, Randi 😀

[@pranavamali05]

Thnku

[@PreMath]

You are very welcome. Thank you for your feedback! Cheers! You are awesome, Pranav 😀

[@asalamkamal6365]

Great!

[@sameerqureshi-kh7cc]

That's our Premath the Rocker 😊👍

[@PreMath]

So nice of you. Good to hear from you, Sameer Qureshi. Thank you for your feedback! Cheers! You are the best 😀 Love and prayers from the USA!

[@janashanthan3210]

Good video

[@ramanivenkata3161]

Though a simple sum, excellent thinking in splitting the last figure of 71889 to 71888+1

[@xyz9250]

From the first two equations one can get ( 9^x)^y=6 or 9^xy = 6, take log_3 , log_3(9^xy) = log_3(6) log_3(3^2xy) = log_3(2) + log_3(3) 2xy = log_3(2) + 1 log_3(2) = 2xy -1

[@VipinKumar-xy9ls]

Alternative- 1. Replace 11^y by 9^xy 2. Use property if a^b = x then logx at base a is equal to b 3. Use property logx at base y is equal to logx/logy 4. Use log6 = log(2*3)=log2 + log3 and log9=log(3*3)=log3 + log3 5. Divide Nr and Dr by log3 and solve

[@mengpheapMEF]

Thanks

[@rajendraameta7993]

log 11 = 2 x log 3...from eq.1 & log 11= (1/y)( log2 +log 3).equating both we get 2xy = (log 2+log 3)/log3=1+(log2/log3) so log2/log3 =2xy-1 & so the ans

[@AmirgabYT2185]

log3(2)=2xy-1

[@georgef9714]

3^(2xy)=11^y=6 take log base 3 on both sides. 2xy*log(3)=log(6)=log(3*2) 2xy*log(3)=log(2)+log(3) 2xy*1=log(2)+1 sooo log(base 3)(2)=2xy-1 problema risolto

[@PreMath]

Thank you for your feedback! Cheers! You are awesome, George 😀

[@baezmarklimuel7871]

Same solution!

[@whycodewhenyoucanai]

For me, the connector is the 11 from both equations. Let's start from the second equation, 11^y = 6. Taking logs base 3, log_3 (11^y) = log_3 (6) = log_3 (3 * 2) = log_3 (3) + log_3 (2) = 1 + log_3 (2) So we have log_3 (11^y) = 1 + log_3 (2) At this point we already have log_3 (2). The left side becomes y * log_3 (11) So if we move y to the right side, we get log_3 (11) = [1 + log_3 (2)] / y Notice that we can get log_3 (11) from the first equation, log_3 (9^x) = log_3 (11) x * log_3 (9) = x * log_3 (3^2) = 2x * log_3 (3) = 2x * 1 Matching the two log_3 (11)s, we get 2x = [1 + log_3 (2)] / y 2xy - 1 = log_3 (2)

[@devondevon4366]

Answer =2xy-1 9^x=11 equation 1 11^y=6 equation 2 11= 6^1/y 9^x = 11 hence 9^x = 6^1/y 9^xy=6 3^2xy= 3 x 2 3^2xy/3 =2 3^2xy-1 =2 or log base 3 2 = 2xy-1 hence answer in term of xy = 2xy -1

[@PreMath]

Very well done and presented! Thank you for your feedback! Cheers! You are the best, Devon 😀

[@zplusacademy5718]

Very nice and beautiful ❤️

[@petrmatejka4900]

Easier (and quicker) would be to use log base 3 from beginning. Instead of using log base 10. From 1st equation you will have 2.x = Log 11 (base 3). From 2nd equation you will have y.log 11 (base 3) = Log 3 + log 2 (both base 3). Multiply 1st by y and finding that left hand of multiplied equation equals right hand of the second you get 2x.y = log 3 + log 2 (both base 3). and since Log 3 (base 3) =1 You will get Log 2 (base 3) = 2xy-1

[@hsduhiufdifeijoiff9558]

Well, If I had solved it, I would've raised the power of 9^x up y. Hence, 11^y=6, then 9^(xy)=6. This can be written as 3^(2xy)=6. From here I can take log_3 : log_3 3^(2xy) = log_3(2*3) So, 2xy = log_3 2 + log_3 3 log_3 2= 2xy-1 :)

[@Anonymous-ek3ud]

Take log on both sides . Then divide the two equations log 11 gets cancelled. Then it will be more easier and it will be shorter approach.

[@PreMath]

Great tip. Many approaches possible. Thank you for your feedback! Cheers! You are awesome 😀

[@sandanadurair5862]

9^x=11 2xlog3=log11........1 11^y= 6 ylog11 = log6 Log11 =log6/y....2 From 1&2 2xlog3 = log6/y 2xy = log6/log3 Log2/log3+log3/log3=2xy Log2/log3 = 2xy-1 Log2 to base 3 =2xy-1

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome 😀

[@omolayoeunice5106]

👍🏻👍🏻👍🏻

[@fitefitefitefite]

I've found a short method. (9^x)^y=6 so, 3^2xy=6. Then 2xylog3=log3+log2, and 2xy=(log3+log2)/log3. Finally 2xy-1=log2/log3

[@PreMath]

Many approaches possible. Thank you for your feedback! Cheers! You are awesome 😀

[@fitefitefitefite]

@@PreMath Your channel is addictive. Thank you very much for helping us. Hugs from Cádiz (Spain)

[@PreMath]

@@fitefitefitefite You are the best. Love and prayers from the USA!

[@TheLukeLsd]

I used this steps: I) xlog(9)=log(11); ylog(11)=log(6); II) xlog(9)=xlog(3²)=2xlog(3)=log(11); Substituting log(11) ylog(11)=y(2xlog(3))=2xylog(3)= log(6) => 2xy = log(6)/log(3)=log3(6)=log3(3)+log3(2)= 1+log3(2) => 2xy=1+log3(2)=>2xy-1=log3(2)

[@mustafizrahman2822]

Very simple. The answer will 2xy-1

[@PreMath]

Great, Mustafiz Thank you for your feedback! Cheers! You are awesome 😀

[@aryamansingh4043]

I do it easily

[@AlexandraMarchenkova]

2xy-1 ✅

[@aydnyesildal4145]

This could be solved as 3^a=2 what is a? 3^(2xy-1) = 2 so a is 2xy-1

[@PreMath]

Thank you for your feedback! Cheers! You are awesome 😀

[@arregull]

9^x = 11 11^y = 6 Therefore: 9^xy = 6 3^2xy = 2*3 3^2xy / 3 = 2 3^(2xy-1) =2 Convert to log. 2xy-1.

[@PreMath]

Many approaches possible. Thank you for your feedback! Cheers! You are awesome 😀

[@rustemomeroruc5977]

log 11 where

[@owlsmath]

Like it! Thanks for video! One thing i noticed is a problem like this sometimes asks for you to find x*y. So i did that even though the problem doesn't call for it and then you kind of luck into the answer that way! 🤣

[@ganeshdas3174]

In terms of xy the value is = 2xy - 1

[@josephoyek6574]

Me: *sees thumbnail _HOW THE FVCK!!_ "...in terms of x and y" Me: Oh wait, I see.

[@lemonwong1873]

ANS IS ( " xy - 1/2 ")

[@JSSTyger]

2xy-1

[@hans-rudigerdrzimmermann]

Comments or critics on the solution video by this channel creator are leading to NO HEART from yourside. Pls think about it and change!!!

[@devondevon4366]

answer 2xy-1

[@PreMath]

Super

[@حمیدرضارحمانیراد]

I have another solution We can multiply x and y after take log on 9^x and 11^y That's faster Try it!

[@sudiptaroy2564]

Could be Solved more easily and simply.

[@aryamansingh4043]

But my answer is y-1

[@om8692]

I have done this with much more less steps

[@PreMath]

Great, Pratap Many approaches possible. Thank you for your feedback! Cheers! You are awesome 😀

[@indrajitkatira5953]

Your method is v lenthy. Write as x=log11with base 9, y=log6to base 11. Now eliminate log11, simplify we get answer 2xy-1

[@jonpopelka]

r=√(csc(2θ)*log3(2)+1) haha

[@tathagatamallick7200]

Too much lengthy solution.

[@susennath6035]

Monotonous math

[@PreMath]

Thank you Susen for your feedback! Cheers! You are awesome 😀