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"I chose a = 5 and b = 3 because the equation has the form 'a^n - b^n = k', where k is a constant. This form is reminiscent of the difference of powers formula, which is often used in algebra. By choosing a and b to be integers, I was able to simplify the equation and test possible values of n. Sorry for late reply
The Cauchy-Schwarz inequality states that for any real numbers a,b,c, and d (a^2+b^2)(c^2+d^2)>=(ac+bd)^2 Using the given values: (4)(9)≥(ac+bd)^2 36≥(ac+bd)^2 Taking the square root of both sides: 6≥∣ac+bd∣ Thus, the maximum value of ac+bd is 6 This approach uses basic algebra and the concept of inequalities .
@@DommarajuJyothi I expected this answer (a^2+b^2)(c^2+d^2)+2abcd-2abcd=4×9=36 a^2c^2+b^2d^2+2ac×bd+a^2d^2+b^2c^2-2ad×bc =(ac+bd)^2+(ad-bc)^2=36 Max value (ac+bd) is √36=6 by taking ad-bc=0.
At 3.15 we see a case of formal fallacy. So if we put x=1, it would lead to unequal terms on both sides does that mean that positive values would also not satisfy the equation. And we should neither take positive values nor negative values. So its not that negative values are not satisfying the equation. Its X=-1, which is not satisfying the equation. Conclusion is correct though we should not take the negative values.
"Haha, nice catch! You're right, if all questions are Olympiad questions, it's a bit of a brain twister! But let's flip it around - if we make every question a compelling one, like a clickbait headline, we'll attract more viewers and make math more engaging!"
@@DommarajuJyothi thank you.I taught my grand daughter (grade 5) by this method using LCM(coefficients of x,y) to get other solutions 81(-38)+LCM(81,237)k+237×13-LCM(81,237)k where k is an integer.
"That's amazing! I love seeing innovative teaching methods like this. You're not only sharing your math expertise with your granddaughter, but also inspiring her to think creatively. Keep up the fantastic work, and I'm sure she'll become a math whiz in no time!"
"'Mod' and 'equality' aren't exactly interchangeable, but in modular arithmetic, a congruence relation can imply equality in certain cases. Think of it as a 'conditional equality' that depends on the modulus. In my solution, I used congruence to imply equality.
Logarithm Property You Should Know: a^√(logb/loga) = b^√(loga/logb) Prove: a^√(logb/loga) = b^√(loga/logb) Let: u = a^√(logb/loga), v = b^√(loga/logb) logu = log[a^√(logb/loga)] = [√(logb/loga)](√loga)² = √[(logb)(loga)] logv = log[b^√(loga/logb)] = [√(loga/logb)](√logb)² = √[(loga)(logb)] logu = √[(logb)(loga)] = √[(loga)(logb)] = logv u = v; a^√(logb/loga) = b^√(loga/logb) Note: The completed set of character format for “logₐb” is not available online
Here's a much simpler solution: recall log(2a) = ln(2a) / ln(10) a^(ln(2a) / ln(10)) = 5 take natural log of both sides: (ln(2a)./ ln(10)).ln(a) = ln(5) so, ln(2a).ln(a) = ln(5).ln(10) by inspection a = 5 (since also 2a = 10) BUT recall ln(a).ln(b) = ln(a^-1).ln(b^-1) = ln(1/a).ln(1/b) so there is another solution where a = 1/a so, 1/2a = 5 and 1/a = 10, so a = 1/10
"Thank you for pointing that out!I'll make sure to correct the thumbnail to reflect the correct operation (plus instead of minus). I appreciate your feedback. Thank you so... much
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!
"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!"
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.
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!
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
