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I did it in my head.

Sqrt[3x+Sqrt[4x]]=x x=0 x=4

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Brazil Math Olympiad: 8^logx - 2^logx = 5!; x = ? x > 0, x is not a complex value root 8^logx - 2^logx = 2^3logx - 2^logx = (2^logx)³ - 2^logx = 120 Let: y = 2^logx; (2^logx)³ - 2^logx = y³ - y = 120, y³ - y - 120 = 0 y³ - 125 - y + 5 = 0, (y³ - 5³) - (y - 5) = (y - 5)(y² + y + 25 - 1) = 0 (y - 5)(y² + y + 24) = 0, y = 2^logx > 0, y² + y + 24 > 0; y - 5 = 0, y = 5 y = 2^logx = 5, log(2^logx) = log5, logx = log₂5= 2.322; x = 10²·³²² Answer check: x = 10²·³²² = 209.894: logx = 2.322 8^logx - 2^logx = 8²·³²² - 2²·³²² = 120 = 5!; Confirmed The calculation was achieved on a smartphone with a standard calculator app Final answer: x = 10²·³²² = 209.894

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5^(Log[2,10])≈209.85923958666630317009678021878406105116185027138480372706929130350846724744115796017140141837433884240451544038441839369207511371129111702474699933960660643784773376084089463204952021758234719668404358826167293500606911551444117786898760982244401165358056339313778365096039167025625158280914139352884146281460364467749300195370770646671418317034941661610608867519252480785132900675030681979670289966779446483003465458591603287898735881661291219772919423918172843500818309430565657090485875738058886684579214785461972609388255827389101733414983549590216258091619555955947697035081595266761158318701772083

X=5^(Log[2,1.25]+3)

Input 8^log_10(5^(log(2, 1.25) + 3)) - 2^log_10(5^(log(2, 1.25) + 3)) = 5! Result True

No problemat all 3!5!7!=n! there is no 11 factor so n<11 thERE IS one more "5" PRIME FACTOR SO THE ONLY POSIBILITY IS n=10 it is enough to check.

2^logx = u logxlog2 = logu logx = log₂u => x = 10^(log₂u) u³ - u = 5! = 120 = 5³ - 5 u³ - 5³ - (u - 5) = 0 (u - 5)(u² + 5u + 24) = 0 u - 5 = 0 => u = 5 *x = 10^(log₂5)* u² + 5u + 24 = 0 => no real solutions

(1 + 2 m) ( 1 + 2 n) = 2 * 38 + 1 Case I : 1 + 2 m = 77, 1 + 2 n = 1 Hereby m = 38, n = 0 Case II: 1 + 2 m = 11, 1 + 2 n = 7 Hereby m = 5, n = 3 Case III: 1 + 2 m = 7, 1 + 2 n = 11 Hereby m = 3, n = 5 Case IV: 1 + 2 m = 1, 1 + 2 n = 77 Hereby m = 0, n = 38

Got it . Very easy. 😊😊😊

Решается за 30 секунд. 2 - очевидный корень, у легко доказать, что уравнение имеет 1 корень.

This is too long and your time will run out! 😁

that is the most 😢 😮 I have ever seen.!!??

If 3!×5!×7! is a factorial of a certain number, the number is necessarily larger than 7 and smaller than 11 that is a prime next to 7. And the factorial divided by 7! necessarily includes a prime factor 5. Therefore, m-5=10 is necessary. What we have to do is confirming the sufficiency.

A=3^（X-4），B=3^(Y-4), A+B=244, A*B=3^(X-4+Y-4)=3^5=243, A B is roots of T^2-244T+243=0, T=(1,243), (A,B)=(1,243),(243,1), (X,Y)=(4,9),(9,4)

How is that simpler?

Very interesting and instructive video. Dear Professor, can you make others with factorial calculation?

Very good question 👍

Nice video man. Keep it up

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Let K=（A+B）/（A-B）， A/B=（K+1）/（K-1）， （K+1）^4+（K-1）^4=10（K+1）^2*（K-1）^2， 2（K^4+6K^2+1）=10（K^4-2K^2+1）， 8K^4-32K^2+8=0， K^4-4K^2+1=0，

これは凄い！

こんなの暗算で、３と−３とわかるよね。計算は必要ない。

Fraco roblema

Math Olympiad Exponential Equation: 5¹⁰ˣ = x²; x = ? (5¹⁰ˣ)¹⸍²ˣ = (x²)¹⸍²ˣ, x¹⸍ˣ = 5⁵ = [1/(1/5)]¹⸍⁽¹⸍⁵⁾ = [(- 1/5)⁻¹]¹⸍⁽¹⸍⁵⁾ = (- 1/5)⁻¹⸍⁽¹⸍⁵⁾ = (- 1/5)¹⸍⁽⁻¹⸍⁵⁾; x = - 1/5 Answer check: x = - 1/5: 5¹⁰ˣ = 5¹⁰⁽⁻¹⸍⁵⁾ = 5⁻² = 1/25 = (- 1/5)² = x²; Confirmed Final answer: x = - 1/5

You take a complicated problem ensuring that the most basic math students can’t follow the logic. Then you take a bunch of steps to find sqrt(-32) for more sophisticated students A while back you did the value of a quadratic equation solved for (1/x) and with no discussion showed that to solve for X just put 2c in the denominator That would have been nice to show that proof

Square Root Simplification: x = √[5x + (√6x)]; x = ? x² = 5x + (√6x), (√x)⁴ - 5(√x)² - (√6x) = 0, Let: y = √x (√x)⁴ - 5(√x)² - (√6x) = y⁴ - 5y² - (√6)y = y(y³ - 5y - √6) = 0 y³ - 5y - √6 = [y³ - (√6)³] - (5y - 5√6) = (y - √6)[y² + (√6)y + 6 - 5] y⁴ - 5y² - (√6) = y(y³ - 5y - √6) = y(y - √6)[y² + (√6)y + √6 - 1] = 0 y = 0, y - √6 = 0; y = √6 or y² + (√6)y + 1 = 0, y = (- √6 ± √2)/2; y = √x √x = 0, x = 0; √x = √6, x = 6 or √x = (- √6 ± √2)/2, x = [(- √6 ± √2)/2]² x = (6 + 2 ± 2√12)/4 = (8 ± 4√3)/4 = 2 ± √3 Answer check: x = 0: √[5x + (√6x)] = √(0 + √0) = 0 = x; Confirmed x = 6: √(30 + √36) = √(30 + 6) = √36 = 6 = x; Confirmed x = 2 ± √3: √x = (- √6 ± √2)/2, (√6x) = (√6)(√x) √[5(2 ± √3) + (√6)(- √6 ± √2)/2] = √(7 ± 6√3) ≠ 2 ± √3 = x; Failed Final answer: x = 0 or x = 6

bad method

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let a =2^11, note (a^4-1) =(a-1)*(1+a+a^2+a^3) => n=a+a^2+a^3 = (a^4-1)/(a-1) -1 = a*(a^3-1)/(a-1) => n/(a^3-1) = a/(a-1).

(x/9)^(x² - 1) = 1 x/9 = 1 => *x = 9* (x² - 1 = 0) and (x ≠ 0) x² = 1 => *x = ± 1* x/9 = -1 and (x² - 1) is even *x = -9* => x² - 1 = 80 (even)

27 3^3 (a ➖ 3a+3),

Math Olympiad Algebra: (x - 5)!(y - 2)! = 10!; x + y = ? 10! = (10)(9)(8)7! = (720)7! = (6!)(7!) (x - 5)!(y - 2)! = (0!)(10!) = (10!)(0!) = (1)(10!) = (10!)(1) = (6!)(7!) = (7!)(6!) 1. (x - 5)! = 0! = (5 - 5)!, (y - 2)! = 10! = (12 - 2)!; x = 5, y = 12 (x - 5)! = 10! = (15 - 5)!; (y - 2)! = 0! = (2 - 2)!; x = 15, y = 2 2. (x - 5)! = 1 = 6 - 5, (y - 2)! = 10! = (12 - 2)!; x = 6, y = 12 (x - 5)! = 10! = (15 - 5)!; (y - 2)! = 1 = 3 - 2; x = 15, y = 3 3. (x - 5)! = 6! = (11 - 5)!; (y - 2)! = 7! = (9 - 2)!; x = 11, y = 9 (x - 5)! = 7! = (12 - 5)!, (y - 2)! = 6! = (8 - 2)!; x = 12, y = 8 x + y = 5 + 12 = 15 + 2 =17; x + y = 6 + 12 = 5 + 13 = 18 or x + y = 11 + 9 = 12 + 8 = 20 Answer check: x = 5, y = 12; x = 15, y = 2 (x - 5)!(y - 2)! = (5 - 5)!(12 - 2)! = (0!)(10!) = (1)(10!) = 10!; Confirmed (15 - 5)!(2 - 2)! = (10!)(0!) = 10!; Confirmed x = 6, y = 12; x = 15, y = 3 (6 - 5)!(12 - 2)! = (1!)(10!) = 10!; Confirmed (15 - 5)!(3 - 2)! = (10!)(1) = 10!; Confirmed x = 11, y = 9; x = 12, y = 8 (11 - 5)!(9 - 2)! = (6!)(7!) = (10)(9)(8)7! = 10!; Confirmed (12 - 5)!(8 - 2)! = (7!)(6!) = (6!)(7!) = (10)(9)(8)7! = 10!; Confirmed Final answer: x + y = 17, x + y = 18 or x + y = 20

(x^2-1)*log(x)=(x^2-1)*log(9) this equation is wrong when x=-9

Four obvious solutions: -9, -1, 1, 9

(x/9) ^ (x² - 1) = 1 1. x/9 = 1 => x = 9 2. x/9 = -1 if x² - 1 is even x = -9 , and (-9)² - 1 = 80 is even 3. x² - 1 = 0 => x = ±1 Ans: x = ±1, ±9

Hermoso ejercicio. Sí, la Matemática es hermosas..

divide powers by 45*x^2 => 243^(1/45) = x^(1/x^2) , 243 =3^5 => 3^(5/45) = 3^(1/9) = x^(1/x^2) =>x=3

Molto originale!!!

Came here from IG. Most definitely glad I did. Great song with an even better lyrics🙏🏾

Your video was a game-changer for me. Thank you! 😍

very very nice 🤩

y = 35/x x² - y² = 24 x² - (35/x)² = 24 x⁴ - 24x² - 35² = 0 x⁴ - 24x² - 5²*7² = 0 (x² + 5²) (x² - 7²) = 0 x² = -5²(undefined), 7² 1. x = 7, y = 35/7 = 5 ==> x + y = 12 2. x = -7, y = 35/(-7) = -5 ==> x + y = -12 Ans: x + y = ±12

5^(10x) = x² 5^(5x) = ± x 5^(5x) = x 5xln5 = lnx (1/x)lnx = 5ln5 (1/x)ln(1/x) = -5ln5 ln(1/x) = W(-5ln5) *x = e^-W(-5ln5)* 5^(5x) = -x u = -x => x = -u 5^(-5u) = u (-5u)ln5 = lnu (1/u)lnu = -5ln5 (1/u)ln(1/u) = 5ln5 ln(1/u) = ln5 u = 1/5 => *x = -1/5*

Thanks also for those with alternative solutions ❤

Cant we just wite √8 as 2√2 instead of lengthy calculations. Make maths simple and mot boring. However you presentation is very good. Thanks

Hi, Thanks for your interesting problem. Here is the way I solved it. Of course I didn't look at your solution. Tell me if you like my solution. Greetings and keep up the good work. Recall. If a^4+b^4=10a^2.b^2 then calculat (a+b)/(a-b)=... Begin Let u=(a+b)/(a-b) Then u is not defined for a=b (so also for a=b=0) Then for the value u, a<>0 and b<>0 and a<>b Let's assume that a,b € R Let's make the calculations below First we can write, a^4+b^4=(a^2+b^2)^2-2a^2.b^2=10a^2.b^2 => (a^2+b^2)^2-2a^2.b^2=10a^2.b^2 (i) => (a^2+b^2)^2=12a^2.b^2 => (a^2+b^2)=+sqrt(12)a.b or (a^2+b^2)=-sqrt(12)a.b => (a+b)^2-2ab=+sqrt(12)a.b or (a+b)^2-2ab=-sqrt(12)a.b => (a+b)^2=(2+sqrt(12))a.b or (a+b)^2=(2-sqrt(12))a.b As a,b € R and a > b and sqrt(12)>2 then Case (i.i) (a+b)^2=(2+sqrt(12))a.b on the condition that a and b has got the same sign or Case (i.ii) (a+b)^2=(2-sqrt(12))a.b on the condition that a and b has got the opposite sign Second we can write a^4+b^4=(a^2-b^2)^2+2a^2.b^2=10a^2.b^2 => (a^2-b^2)^2+2a^2.b^2=10a^2.b^2 (ii) => (a^2-b^2)^2=8a^2.b^2 => (a^2-b^2)=+sqrt(8)a.b or (a^2-b^2)=-sqrt(8)a.b => (a+b)(a-b)=+sqrt(8)a.b or (a+b)(a-b)=-sqrt(8)a.b By assuming that a > b As a,b € R then Case (ii.i) (a+b)(a-b)=+sqrt(8)a.b on the condition that a and b has got the same sign Case (ii.ii) (a+b)(a-b)=-sqrt(8)a.b on the condition that a and b has got the opposite sign Still assuming a > b, if we remark that (a+b)^2/[(a+b)(a-b)]=(a+b)/(a-b)=u then we can calculate the requested value on the considered two sub condition cases. Condition i: a and b has got the same sign (and a > b) then u=(a+b)/(a-b)=[(2+sqrt(12))a.b]/[+sqrt(8)a.b]=(2+sqrt(12))/sqrt(8) u=(a+b)/(a-b)=(2+2sqrt(3))/2sqrt(2)=(1+sqrt(3))/sqrt(2)=(sqrt(2)+sqrt(3))/2 Or Condition ii: a and b has got the opposite sign (and a > b) then u=(a+b)/(a-b)=[(2-sqrt(12))a.b]/[-sqrt(8)a.b]=(-2+sqrt(12))/sqrt(8) u=(a+b)/(a-b)=(-2+2sqrt(3))/2sqrt(2)=(-1+sqrt(3))/sqrt(2)=(-sqrt(2)+sqrt(3))/2 Newly assuming that b > a As a,b € R then Case (ii.iii) (a+b)(a-b)=+sqrt(8)a.b on the condition that a and b has got the opposite sign Case (ii.iv) (a+b)(a-b)=-sqrt(8)a.b on the condition that a and b has got the same sign Newly assuming b > a, if we remark that (a+b)^2/[(a+b)(a-b)]=(a+b)/(a-b)=u then we can calculate the requested value on the considered two sub condition cases. Condition iii: a and b has got the same sign (and b > a) then u=(a+b)/(a-b)=[(2+sqrt(12))a.b]/[-sqrt(8)a.b]=-(2+sqrt(12))/sqrt(8) u=(a+b)/(a-b)=-(2+2sqrt(3))/2sqrt(2)=-(1+sqrt(3))/sqrt(2)=-(sqrt(2)+sqrt(3))/2 Or Condition iv: a and b has got the opposite sign (and b > a) then u=(a+b)/(a-b)=[(2-sqrt(12))a.b]/[+sqrt(8)a.b]=(2-sqrt(12))/sqrt(8) u=(a+b)/(a-b)=(2-2sqrt(3))/2sqrt(2)=(1-sqrt(3))/sqrt(2)=(sqrt(2)-sqrt(3))/2 To conclude, on the conditions that a,b are real numbers condition i : a>b and a and b has got the same sign => u=(a+b)/(a-b)=(+sqrt(2)+sqrt(3))/2 condition ii : a>b and a and b has got opposite sign => u=(a+b)/(a-b)=(-sqrt(2)+sqrt(3))/2 condition iii: a<b and a and b has got the same sign => u=(a+b)/(a-b)=(-sqrt(2)-sqrt(3))/2 condition iv : a<b and a and b has got opposite sign => u=(a+b)/(a-b)=(+sqrt(2)-sqrt(3))/2 END

Using logarithm table its very eazy