2 сағат бұрын
12 сағат бұрын
19 сағат бұрын
Күн бұрын
21 күн бұрын
Пікірлер
@andrea-mj9ce 8 сағат бұрын
The thumbnail is not the same as the problem of the video
@sudhakarpadmaraju7592 Күн бұрын
Great job
@DommarajuJyothi Күн бұрын
🙏
@walterwen2975 2 күн бұрын
Hard Olympiad and Log exponential: a^log(2a) = 5; a = ? log[a^log(2a)] = log5, [log(2a)](loga) = (log2 + loga)(loga) = log(10/2) (loga)² + (log2)(loga) - (log10 - log2) = (loga)² + (log2)(loga) - (1 - log2) = 0 (loga)² + (log2)(loga) + (log2 - 1) = [loga + (log2 - 1)](loga + 1) = 0 loga + log2 - 1 = 0; loga = 1 - log2 or loga + 1 = 0; loga = - 1 loga = 1 - log2 = log(10/2) = log5, a = 5; loga = - 1 = log(1/10), a = 1/10 Answer check: a = 5: a^log(2a) = 5^log[(2)(5)] = 5^log10 = 5¹ = 5; Confirmed a = 1/10: log[a^log(2a)] = [log(1/5)][log(1/10)] = (log5⁻¹)(log10⁻¹) log[a^log(2a)] = (- log5)(- 1) = log5, a^log(2a) = 5; Confirmed Final answer: a = 5 or a = 1/10
@alexeygourevich6967 3 күн бұрын
It's simple, the solution is here. --------------------------------------------------------------------- Logarithmizing: log(2a) * log(a) = log(5). (log(2) + log(a)) * log(a) = log(5). Setting x = log(a): x^2 + x*log(2) - log(5) = 0. Let’s now set c=log(2). Then x^2 + cx - (1-c) = 0. Discriminant D = c^2 + 4(1-c) = c^2 - 4c + 4 = (2-c)^2, so x1 = (-c - (2-c))/2 = -1 => log(a)=-1 => a=10^(-1), x2 = (-c + (2-c))/2 = 1-c => log(a)=1-log(2) => log(a) = log(5) => a=5. So we have two solutions: 1) a=5: it’s THE solution, because 5^(log(2*5)) = 5^(log(10)) = 5^1 = 5. 2) a=10^(-1): (10^(-1))^log(1/5)) = 10^(-(-log(5))) = 10^(log(5)) = 5. It’s THE solution, too! Solution: a=0.1 or a=5. P.S. I just don't think the calculations from 07:50 to 09:50 are very helpful in this problem. But to know approx. values of log(1)...log(10) these, sure, are a very nice trick!
@randomperson5875 3 күн бұрын
what kind of olympiad has these kinds of simple problems? LMAO; I'd like to apply
@DommarajuJyothi 3 күн бұрын
"Haha, fair point! The problem I solved earlier is indeed a simple one. However, Olympic math competitions, like the International Mathematical Olympiad (IMO), feature a range of questions, from basic to advanced. They test problem-solving skills, mathematical thinking, and creativity. If you're interested in exploring math Olympiads, I encourage you to check out resources like the IMO website, Art of Problem Solving, or (link unavailable) Who knows, you might just find yourself representing your country in a future math Olympiad!"
@sudhakarpadmaraju7592 5 күн бұрын
Well said
@DommarajuJyothi 5 күн бұрын
Appreciate your support !
@sudhakarpadmaraju7592 7 күн бұрын
Great Job.❤
@DommarajuJyothi 7 күн бұрын
Thank you! 😊
@user-ee7nw2rx9s 11 күн бұрын
Это уже третий человек который решает этот пример
@DommarajuJyothi 10 күн бұрын
👍
@sudhakarpadmaraju7592 15 күн бұрын
Great work🎉
@DommarajuJyothi 15 күн бұрын
Thank you 🙏
@sudhakarpadmaraju7592 21 күн бұрын
Well said🎉
@DommarajuJyothi 21 күн бұрын
Thank you
@herbertwandha6110 23 күн бұрын
Firstly,I commend you for this good job you do teacher to many out there. In the resulting cubic equation ,the splitting of term + a in terms of -36a and -37a in relationship to the constant term -222 may seem challenging to many students. Suggestion, to consider a cube closest to the constant term as (-216 -6 = -222 ) and associating -222 with a-cubed then factories from there. To sum up, in most cases, the constant term may lead you to early results. Another thing is that your solution in the form: X = log 6 / log 222 , is just good enough...........thanks.
@herbertwandha6110 23 күн бұрын
CORRECTION: Associating -216 with a - cubed.
@DommarajuJyothi 22 күн бұрын
Ok
@herbertwandha6110 23 күн бұрын
In other words, the step : log(37) is not equal to log(36+1).Even log 4 does not mean log (2+2),but could mean log(2*2)=log2 +log2 Remember, you are on the world stage !.....blessings!
@DommarajuJyothi 22 күн бұрын
I apologize for the error. I'll make sure to double check my equations in the future
@sudhakarpadmaraju7592 25 күн бұрын
Good work🎉
@DommarajuJyothi 25 күн бұрын
Thank you! 😊
@user-gd6ck8cr2p 26 күн бұрын
You' re great, please have a look at this: lnx^(cube root) +lnx=10.
@DommarajuJyothi 26 күн бұрын
log x^(cube root) = log x^(1/3) = 1/3 log x 1/3 log x + log x = 10 ( base 10) 4/3 log x = 10 ( base 10) log x = 10 × 3/4 =15/2 ( base 10) log (base 10) of b = a then 10^ a = b. [ exponential logarithm formula) Therefore x = 10 ^ 15/2 x = 10^ 7.5 = 10^ ( 7+ 0.5) x = 10^7 ×10^ 1/2 x = 10^7 × 3.16227766 x = 31622.7766
@user-gd6ck8cr2p 26 күн бұрын
@@DommarajuJyothi very sorry, my fault, if f(x)=lnx, then f(x)^cube root +f(x)=10. {cube root √lnx +lnx=10, x=? , I know only the answer e to the 8}
@DommarajuJyothi 26 күн бұрын
( log x)^1/3 + logx = 10 Let a = log x a^ 1/3 + a = 10 a = 8 log x = 8 x = e^8
@user-gd6ck8cr2p 25 күн бұрын
@@DommarajuJyothi 👍👍, EXCELLENT, may I ask you, which country are you from?
@DommarajuJyothi 25 күн бұрын
I am from India 🇮🇳
@user-qy8re3yx3d 27 күн бұрын
Очевидно, что а=5.
@libardouribe7617 28 күн бұрын
👍
@ranaur1 Ай бұрын
You can also solve it using group theory properties. a^b mod c will always generate a cyclic group of order less than or equal to c. Also a^b mod c = (a^(b-1) mod c).a mod c. So its easier to calculate 5555*4mod 7 than 5555^2 mod 7. So, let' find the sequence 5555 mod 7 = 4 5555^2 mod 7 = 5555*4 mod 7 = 2. 5555^3 = 1. The cyclic group is (4,2,1) of order 3. 2222 mod 3= 2, so 5555^2222 mod 7 is the second element os the group = 2. Using the same reasoning 2222^5555 mod 7 has a group of order 5 and the 5 th element is 5. 2+5 mod 7 is, of course 0.
@DommarajuJyothi Ай бұрын
Absolutely right! Great job spotting that
@UKPEINDANIELU. Ай бұрын
Nice logarithm
@DommarajuJyothi Ай бұрын
Thanks for the feedback! Logarithms can be tricky but they're also really interesting .Do you have a favorite math concept or topic you'd like to see covered next?
@DommarajuJyothi Ай бұрын
Could you please subscribe my channel.
@UKPEINDANIELU. Ай бұрын
@@DommarajuJyothi l need such logarithm questions
@DommarajuJyothi Ай бұрын
Today's post is just what you needed! I posted another logarithm question, check it out and let me know what you think
@user-gd6ck8cr2p 28 күн бұрын
Can you solve please: (49 to the logx) +7=8.(x to the log7)?
@BaskaranRamachandran-fw1ce Ай бұрын
log(a+b) is not equal to log a+log b
@DommarajuJyothi Ай бұрын
"Hi sir! Thank you for pointing out the mistake. You are absolutely right, the equation log(a + b) = log(a) + log(b) is not correct in general. I apologize for the error. The correct rule is log(a × b) = log(a) + log(b), known as the product rule of logarithms. I'll make sure to double-check my equations in the future. Thank you for helping me improve my accuracy!" Please subscribe my channel. Next time I will not disappoint you sir.
@sudhakarpadmaraju7592 Ай бұрын
Good Job
@forgottenfamily Ай бұрын
Before watching the video... A(n) = sum (x=0 -> n-1)(x*x!) 100! ? A(100) BUT n! = n*n-1! = (1 + n-1)*(n-1)! = (n-1)! + (n-1)*(n-1)! subtract (n-1)*(n-1)! from both sides n-1! ? sum(x = 0 -> n-2)(x*x!) = A(n-1) We can repeat this until we get to an easily calculable case 2! ? A(2): 2! ? 1*1 2! > 1*1 Therefore 3! > A(3) and so on until we get to 100! > A(100) (Watch) Well, that's a more precise solution
@DommarajuJyothi Ай бұрын
Thank you for sharing a more formal and precise proof using mathematical induction! Your approach is absolutely correct and provides a stronger foundation for the conclusion. By the recursively applying the formula and simplifying, you've elegantly shown that n!>A(n) for all n, including the specific case of 100! > A(100). Your input is invaluable. I appreciate the time you took to share your insight. This will certainly help others understand the problem more deeply. I hope you like the another concept that cubing of unique pairs 11^3, 22^3,... .If you like this content please subscribe my channel and we'll learn together.
@prach2 Ай бұрын
X=1 ... by guessing.
@TECH-fk8to Ай бұрын
👍
@DommarajuJyothi Ай бұрын
Thank you
@sudhakarpadmaraju7592 Ай бұрын
Great Job 👍
@DommarajuJyothi Ай бұрын
Thanks! 👍
@TECH-fk8to Ай бұрын
👍
@DommarajuJyothi Ай бұрын
Thank you
@TECH-fk8to Ай бұрын
👍
@DommarajuJyothi Ай бұрын
Thank you
@sudhakarpadmaraju7592 2 ай бұрын
Great job
@DommarajuJyothi 2 ай бұрын
Thank you
@Roham_planet__explains 2 ай бұрын
Hello
@DommarajuJyothi 2 ай бұрын
Hi, how are you
@TECH-fk8to 2 ай бұрын
Nice
@DommarajuJyothi 2 ай бұрын
Thanks
@TECH-fk8to 2 ай бұрын
👍
@unflexian 2 ай бұрын
thumbnail doesnt say its integer equation :(
@DommarajuJyothi 2 ай бұрын
Thankyou for your observation! While I didn't explicitly mention in the thumbnail that a &b are integers, I made sure to explain it during the video to ensure clarity for everyone. I appreciate your attention to detail and hope you found the explanation helpful!
@unflexian 2 ай бұрын
@@DommarajuJyothi yeah i saw, great videos from you ! keep up as long as you enjoy it:)
@DommarajuJyothi 2 ай бұрын
Thank you.
@anachronicity 2 ай бұрын
Nice
@DommarajuJyothi 2 ай бұрын
" Glad you liked it!" Thanks for the encouragement
@pritamroy9320 2 ай бұрын
That's a nice solution. Rigours from top to bottom. Only one input I'll give is that from the line (1+2a)(1+2b) = 17, it is more general to say that (1+2a) = either (1 or 17 ) or (-1 or -17) and consequently (1+2b) = either (17 or 1) or (-17 or -1 ). Since we are adding (1+2a) and (1+2b) to find (a+b) , it doesn't matter which one of them is 1 or 17 OR -1 or -17, the sum will be either 18 or -18. So in the end everything checks out.
@DommarajuJyothi 2 ай бұрын
Thank you for your insightful input! I appreciate your suggestion for considering the general solutions of (1 + 2a) and (1 + 2b).
@legend_legend_legend 2 ай бұрын
or just set a = 0, so a+2ab+b = 0+0+b = 8 b=8 a =0 0+8 = 8
@DommarajuJyothi 2 ай бұрын
Thanks for sharing your approach! While setting a=0 simplifies the equation, it's important to note that it only works for specific cases. Exploring various scenarios,like negative integers, allows for a more comprehensive understanding of the problem. Both methods have their merits in different contexts.
@sudhakarpadmaraju7592 2 ай бұрын
Good work
@DommarajuJyothi 2 ай бұрын
Thanks
@TECH-fk8to 2 ай бұрын
👍
@sudhakarpadmaraju7592 2 ай бұрын
Keep going, great work
@sudhakarpadmaraju7592 2 ай бұрын
Great work, proud of you.
@DommarajuJyothi 2 ай бұрын
Thank you so much 😀
@TECH-fk8to 2 ай бұрын
👍
@TECH-fk8to 2 ай бұрын
👍
@TECH-fk8to 2 ай бұрын
Nice
@DommarajuJyothi 2 ай бұрын
Thanks
@sudhakarpadmaraju7592 2 ай бұрын
Great work
@DommarajuJyothi 2 ай бұрын
Thanks
@charlesokuom8747 3 ай бұрын
Well explained
@DommarajuJyothi 3 ай бұрын
Thank you sir
@TECH-fk8to 4 ай бұрын
👍
@sudhakarpadmaraju7592 4 ай бұрын
Nice👍
@DommarajuJyothi 4 ай бұрын
Thank you
@TECH-fk8to 4 ай бұрын
Nice explanation.
@DommarajuJyothi 4 ай бұрын
Thank you. Please subscribe to my channel
@TECH-fk8to 4 ай бұрын
Ok mam
@TECH-fk8to 4 ай бұрын
Good explanation.
@DommarajuJyothi 4 ай бұрын
😊
@TECH-fk8to 4 ай бұрын
Nice explanation.
@DommarajuJyothi 4 ай бұрын
Thank you.
@Xyz-fu4ns 4 ай бұрын
Nice
@DommarajuJyothi 4 ай бұрын
Thank you
@raulrosu 4 ай бұрын
Nice explanation and clean handwriting!
@DommarajuJyothi 4 ай бұрын
Thank you, I really appreciate you taking time for me to express that. And please subscribe.