Відео

81

Nice work

This isn't faster, this is more complex and confusing...

Good job😂

Exponent puzzle solved: 4ˣ - 4ʸ = 24, 2ˣ⁺ʸ = 35; x - y =? 4ˣ > 4ʸ > 24, x > y > 0; 4ˣ - 4ʸ = (2ˣ)² - (2ʸ)² = (2ˣ + 2ʸ)(2ˣ - 2ʸ) = 24, 2ˣ⁺ʸ = 35 4ˣ - 4ʸ = (2ˣ + 2ʸ)(2ˣ - 2ʸ) = 24(1) = (12)(2) = (8)(3) = (6)(4), 2ˣ + 2ʸ > 2ˣ - 2ʸ > 0 1. 2ˣ + 2ʸ = 24; 2ˣ - 2ʸ = 1: 2ˣ = 25/2, 2ʸ = 23/2; 2ˣ⁺ʸ = (25)(23)/4 > 35; Rejected 2. 2ˣ + 2ʸ = 12; 2ˣ - 2ʸ = 2: 2ˣ = 14/2 = 7, 2ʸ = 10/2 = 5; 2ˣ⁺ʸ = (7)(5) = 35; Proved 3. 2ˣ + 2ʸ = 8; 2ˣ - 2ʸ = 3: 2ˣ = 11/2, 2ʸ = 5/2; 2ˣ⁺ʸ = 55/4 < 35; Rejected 4. 2ˣ + 2ʸ = 6; 2ˣ - 2ʸ = 4: 2ˣ = 10/2 = 5, 2ʸ = 2/2 = 1; 2ˣ⁺ʸ = (5)(1) < 35; Rejected 2ˣ = 7, 2ʸ = 5, 2ˣ⁺ʸ = 35; 2ˣ/2ʸ = 2ˣ⁻ʸ = 7/5, x - y = log₂(7/5) = 0.485 The calculation was achieved on a smartphone with a standard calculator app Answer check: x - y = log₂(7/5) = 0.485, 2ˣ = 7, 2ʸ = 5: 4ˣ - 4ʸ = 24, 2ˣ⁺ʸ = 35; Confirmed as shown Final answer: x - y = log₂(7/5) = 0.485

Nice

Thank you. A really cool solution. ;)

Amazing work.

Nice

Well said.

Thank you so much mam

Good work.

Good job

Bravo, great work behind the screen.

Great work, keep it up.❤

Excellent explanation.

Excellent 🎉

Good

👍

Great work behind to post this, keep it up.

n = 4. It does not have to be complicated; I KNOW IT!!!

How come you suddenly chose (5,3,2) for (a,b,n) when in fact there are many triplets to choose?

5^n=544+(3^n) 5^n>=544 --> n>=4 as 5⁴=625 For n>=4, 3^n>=81 5^n>=625 (5^n)-3^n>=625-81 >=544 As (5^n)-(3^n)=544, x=4 3^n=(5^n)-544 3^n>544 --> n>=6 as 3⁶=729

Middle school children do not know trigonometry, vectors and matrices.4th method is required.

When dividing like bases write down the base and subtract the indices. 9^-18 cube root divide by 3 =9^-6. Simple.

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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.

Great job done, keep it up🎉.

if every question is an Olympiad question, then which questions are not Olympiad questions?

81x+237y=3 27x+79y=1 27x+81y-2y=1 -2y=1-27k=-26,-53,-80, y=13,40,67,.. x=-38,-117,-196,..

Nice solution. Small thing a,b,c. Should be positive integers not real nos.

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Ok, but why exclude : 38 = 1^2+1^2+6^2 --> 11^2+11^2+66^2 = 4598

how could mod and equality be used interchangeably ???

(log(2a))loga = log5 loga = u => a = 10^u u² + ulog2 - log5 = 0 u = [-log2 ± √(log²2 + 4log5)]/2 u = [-log2 ± √(log²2 - 4log2 + 4)]/2 u = [-log2 ± (log2 - 2)]/2 u = -1 => *a = 1/10* u = 1 - log2 = log5 => *a = 5*

Great work behind the screen, keep it up.🎉

Awesome 👌

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

Great work, very well explained.

Awesome! Very helpful. :)

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

There is a sign of negative bit here you expressed it positive Thats quite disappointing

Thumbnail is wrong Please rectify that

Bulgarian Algebraic Olympiad: (8ˣ + 27ˣ)/(12ˣ + 18ˣ) = 7/6; x = ? (8ˣ + 27ˣ)/(12ˣ + 18ˣ) = (2³ˣ + 3³ˣ)/(2²ˣ3ˣ + 2ˣ3²ˣ) = 7/6, Let: u = 2ˣ, v = 3ˣ (2³ˣ + 3³ˣ)/(2²ˣ3ˣ + 2ˣ3²ˣ) = (u³ + v³)/(u²v + uv²) = 7/6 [(u + v)(u² - uv + v²)]/[uv(u + v)] = (u² - uv + v²)/uv = 7/6; u + v ≠ 0 6u² - 6uv + 6v² = 7uv, 6u² - 13uv + 6v² = 0, (3u - 2v)(2u - 3v) = 0 3u - 2v = 0, 3u = 2v; v/u = 3/2 or 2u - 3v = 0, 2u = 3v; v/u = 2/3 v/u = 3ˣ/2ˣ = (3/2)ˣ = (3/2)¹; x = 1 or v/u = (3/2)ˣ = 2/3 = (3/2)⁻¹; x = - 1 Answer check: x = 1: (8ˣ + 27ˣ)/(12ˣ + 18ˣ) = (8 + 27)/(12 + 18) = 35/30 = 7/6; Confirmed x = - 1: (1/8 + 1/27)/(1/12 + 1/18) = [(12)(18)/(8)(27)][(27 + 8)/(18 + 12)] = 35/30 = 7/6; Confirmed Final answer: x = 1 or x = - 1

Alfa=α in Gre 🇬🇷 greetings, 👍👍

The thumbnail is not the same as the problem of the video

Great job

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

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!

what kind of olympiad has these kinds of simple problems? LMAO; I'd like to apply