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Math Olympiad Algebra: 27ˣ + x = 0; x = ? 27ˣ > 0 > x; 27ˣ = - x Trial-and-error math solution: x = 0: 27⁰ = 1 > 0 x = - 1: 1/27 < - (- 1) = 1, 0 > x > - 1; In-between x = - 1/2: 1/√27 = 1/(3√3) < - (- 1/2) = 1/2; 0 > x > - 1/2, Slightly > - 1/2 x = - 1/3: 1/(³√27) = 1/3 = - (- 1/3); Proved Answer check: 27ˣ + x = 27⁻¹⸍³ + (- 1/3) = 1/3 - 1/3 = 0; Confirmed Final answer: x = - 1/3
Es ist tatsächlich eine schwierige Frage
Go through it over and over again till you understand
Brilliant. Thanks for posting this.
Glad you enjoyed it
??????
Rationalize the denominator
According to Mathematica this would be the complete set of solutions: {-1.49664 - 4.95853 I, -1.49664 + 4.95853 I, 4.98055 - 0.0148474 I, 4.98055 + 0.0148474 I, 5.1, 7.} where I is the imaginary unit.
This root was missing: -27.2436654839064
The solution is too long and stupid
You do an excellent job of using algebra to solve this problem. You seemed to be working towards a known answer (you gave no justification for why you made some of the steps). I suspect that this would likely have been solved by "inspired guesswork" rather than the algebra. Something like ... x^(-x^x) => (x^x^x)^-1 = 2^√2 = 2^(2^(1/2)) hints (i.e. I guess) x is a power of 2 and the ^-1 implies a negative power of 2. Clearly 2^-1 won't work so try 2^-2. It works!
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Sir please l don't understand 😢
Please clarify where you don't understand so that I can help
After 2:20 its so clear m=1/4.
Writing same express. hundred times? Why
To emphasis and demonstrate authority over the topic
Breaking 15 into 9+6 is something I never catch onto by myself. Nice one Sensei.
Good stuff
A nice Math Olympiad Algebra: 16ˣ + 44ˣ = 121ˣ; x = ? 16ˣ > 0; (16ˣ + 44ˣ)/(16ˣ) = (121ˣ)/(16ˣ), (121/16)ˣ = (44/16)ˣ + (16/16)ˣ [(11/4)ˣ]² - (11/4)ˣ - 1 = 0; (11/4)ˣ = (1 + √5)/2; (11/4)ˣ > 0 log(11/4)ˣ = log[(1 + √5)/2] x = log[(1 + √5)/2]/log(11/4) = log1.618/log2.75 = 0.476 Answer check: 16ˣ + 44ˣ = 16^0.476 + 44^0.476 = 3.742 + 6.057 = 9.800 121ˣ = 121^0.476 = 9.804; Confirmed The calculation was achieved on a smartphone with a standard calculator app Final answer: x = 0.476
81^81 9^9 9^9 3^2 3^2 3^23^2 1^1 1^1 1^1 3^2 (x ➖ 3x+3)
until 4.25 minutes, we are at the same place
This is not right in exam there is time limit
Overflowing with love!
This solution is incomplete. You made one solution obvious but didn’t prove there are no other roots. X^x is not a monotonous function for all x’s.
9^4^m = 4^9^m (4^m)*log(9) = (9^m)*log(4) (9^m)/(4^m) = log(9)/log(4) (9/4)^m = log(9)/log(4) m*log(9/4) = log(log(9)/log(4)) m = log(log(9)/log(4))/log(9/4) m = 0.56794128396
16 - x^2 = sqrt(16 - x) 256 - 32*x^2 + x^4 = 16 - x x^4 - 32*x^2 + x + 240 = 0 That's solvable now.
Aye yai yai... 6 = 216^(8^(3*(x^-1))) ln(6) = ln(216^(8^(3*(x^-1)))) ln(6) = (8^(3*(x^-1)))*ln(216) 8^(3*(x^-1)) = ln(6)/ln(216) ln(8^(3*(x^-1))) = ln[ln(6)/ln(216)] 3*(x^-1)*ln(8) = ln[ln(6)/ln(216)] x^-1 = ln[ln(6)/ln(216)]/(3*ln(8)) x = 3*ln(8) / ln[ln(6)/ln(216)] This comes out to x = -5.67836778214. Nothing challenging here at all - just a tedious messy bunch of manipulation. Of course, you could shortcut the whole thing by noticing that 216 = 6^3, so you see immediately that 8^(3/x) has to be 1/3. The process above is just what you'd do if you had absolutely no insight into things at all - the "purely mechanical process."
m=½ln(ln2/ln3)/(ln2-ln3)≈0.568 😬
What a monumental waste of time!
It's not a waste of time but a demonstration of authority on the topic
@@superacademy247 Nonsense. You didn't do anything particularly clever, and you didn't arrive at a surprising or interesting answer. It was a wasted effort in which you tediously rewrote expressions rather than simply writing the result of the expression. Your grindingly slow approach to a solution won't inspire anyone.
Try usin expansion (a/b)^x = 1 + xln(a/b) + [x^2ln(a/b)^2]/2 ...... Using this property there must be no real solution
3^2 3^25 8^8 2^3:4^6;3^1 3^80 1^1 1^5 2^3 2^3 1^1 2^2 3^2 1^1 3^2^401^1 1^1 1^1 1^1 1^1 3^2 2^20 3^1 1^5^4 3 5^1 2^2 3 1^1 1^2 32 (x ➖ 3x+2
Math Olympiad Problem: 9^(4^m) = 4^(9^m); m = ? log[9^(4^m)] = log[4^(9^m)], (9^m)log4 = (4^m)log9 (9^m)/(4^m) = log₄9 = 1.585; (9/4)^m = 1.585, 2.25^m = 1.585 log(2.25^m) = log1.585, m = (log1.585)/(log2.25) = 0.200/0.352 = 0.568 Answer check: 9^(4^m) = 9^(4^0.568) = 125.068 4^(9^m) = 4^(9^0.568) = 125.097= 9^(4^m); Confirmed The calculation was achieved on a smartphone with a standard calculator app Final answer: m = 0.568
multiplying both sides by 5, then let y=5x-25. we have (logy)^2=1, ....
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Too much talk and unnecessary writing!
Teach me more on log because I'm going for a math competition you might even coach me.what do you think?
Thanks for your inquiry. Write me email on [email protected]
а^m ??? q^m??? 9^m??? Научитесь писать буквы и цифры!
I'll do more to achieve good handwriting. Thanks for your concern .
(5x)^2(25x)= {25x^2 625 x^2}= 600x (x (minus ➖ 3x+2)
[m^(1/2+1)^1/2+1}^1/2+1]^1/2=m^15/16 =>3^(15/16)×=2^15 =>3^×=2^16 =>×=16log@3 ( 2) =>3^×=2^16
Detailed explanation ! Good !
Glad it was helpful!
W2=W(-ln3/9)=-0,140478921099014 x2=-W2/ln3=0,1278694244985
x=16log⅔
You have to verify that argument of W is not less to -1, which you forgot. Bad!
3^x-2=x^1 ; x-2=1 ; x=1+2 ; x=3
Le soluzioni sono due. Errato
Aklımdan geçeni bul şeklinde bir soru
3 is an obvious solution. One can also take the log base 3 of each side and see the answer almost immediately. More inspiration and less perspiration.
Non risolvi molto facendo il logaritmo di ambo i membri... Quello che è ovvio, è farsi il grafico di ambo i membri e ovviamente si vede subito che ci sono due soluzioni. E sai pure più o meno dove si trovano
How to calc the 2nd solution ?
By plotting the graph you can get the point of intersections. Pretty straight forward. But there's is a special method using Lambert W Function. I'll create a video to demonstrate that
@@superacademy247la funzione di Lambert non sposta il problema... In definitiva la seconda soluzione non è esprimibile in modo esatto. Va trovata con un metodo approssimato
A Nice Math Olympiad Algebra Problem: 3ˣ⁻² = x; x = ? 3ˣ⁻² > 0, x ≥ 0 Solve the exponential equation by Trial-and-error math: 3ˣ⁻² = x x = 0: 3⁻² = 1/9 > x = 0 x = 1: 3⁻¹ = 1/3 < 1; 1 > x > 0, Close to 0 x = 0.2: 3⁻¹ꞏ⁸ = 0.128 < 0.2, 0.2 > x > 0, Slightly < 0.2 x = 0.125: 3⁻¹ꞏ⁸⁷⁵ = 0.127 > 1.125, 0.2 > x > 0.125 x = 0.128: 3⁻¹ꞏ⁸⁷² = 0.128 = 0.128; Proved x = 3: 3¹ = 3; Proved x = 0.128 or x = 3 Answer check: x = 0.128 or x = 3: 3ˣ⁻² = x; Confirmed as shown Final answer: x = 0.128 or x = 3
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Mi dispiace ma la prima soluzione è troppo approssimativa. In questo caso basterebbe fare un grafico di ambo i membri... Ovviamente puoi approssimare in vari modi la soluzione più piccola
The actual values are:2,-2,3,-1
{(3^7)^(1/9)}/3 = 3^(7/9) * 3^(-9/9) = 3^(-2/9) = 1/ (3^2/9) = 1/9√(3^2) = 1 / 9√9 Or simply (1/9)^(1/9) = 9√1 / 9√9 = 1 / 9√9
Nice job !
Thanks!
A Nice Algebra Problem: x + y = 2, xy = 4; x, y = ? (x - y)² = (x + y)² - 4xy = 2² - 4(4) = - 12 = (2i√3)²; x - y = ± 2i√3, x + y = 2 2x = 2 ± 2i√3, x = 1 ± i√3; 2y = 2 -/+ 2i√3, y = 1 -/+ i√3 Answer check: x = 1 ± i√3, y = 1 -/+ i√3 x + y = (1 ± i√3) + (1 -/+ i√3) = 2; Confirmed xy = (1 ± i√3)(1 -/+ i√3) = 1 - (i√3)² = 1 + 3 = 4; Confirmed Final answer: x = 1 + i√3, y = 1 - i√3 or x = 1 - i√3, y = 1 + i√3; Complex value roots
2^6=64 X+1=2 Rest is simpel
ln(x - 5)[ln(5x - 25)]/ln10 = ln2 ln(x - 5) = u u(u + ln5) = (ln2)(ln2 + ln5) u² + uln5 - (ln2)(ln2 + ln5) = 0 u = (-ln5 ± √[ln²5 + 4ln²2 + 4(ln2)ln5]/2 u = [-ln5 ± (ln5 + 2ln2)]/2 u = ln2 => ln(x - 5) = ln2 => *x = 7* u = -ln10 => x - 5 = 1/10 => *x = 5.1*
Use this simple trick: Substitute x=1+z and y=1-z into the second equation and rearrange. (z+1)(z-1)=z^2-1=-4; z^2=-3 and z=±i√3. Thus, x=1+z=1±i√3 and y=1-z=1∓i√3