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Great 👍 sir please voice

Thanks for vedio🎉🎉

Great job 👍 sir you are very talented and hardworking thanks for vedios

3^30*ln3=ln3*(30-x)*e^(ln3*(30-x)) , 3^3*3^27*ln3=ln3*(30-x)*e^(ln3*(30-x)) , 3^3*ln3*e^((ln3)^27)=27*ln3*e^(27*ln3) , 27*ln3=ln3*(30-x) , 27=30-x , x=3 , test , 3^3+3=27+3 , --> 30 , OK ,

x = 3: 3 cubed is 27 and 27 plus 3 is 30. Thank you. May you and yours stay well and prosper.

x = 3: 5 to the power of 3 is 125 and 125 plus 3 is 128. Thanks. May you and yours stay well and prosper.

16=4^2 =>4^2x=4^2 comparison of exponents 2x=2 X=1 much quicker / easier

Sir thanks for this video

by faktoring , 2x^4 - 11x^2 + 5 = 0 , x^2(x^2-1)-5(x^2-1)=0 , (x^2-1)(x^2-5)=0 , x^2-1=0 , x=+/-1/V2 , 2 -1 x^2-5=0 , x%+/-V5 , -10 5 solu , x= 1/V2 , -1/V2 , V5 , -V5 ,

3^x + x = 85 --> 3^x+ x = 81 + 4 --> 3^x + x = 3^4 + 4 --> x = 4

nice video XD

x=1

Sir voice over bi kiya kary sath may be improvement ay gi channel ko grow karny ma 🎀

❤❤❤

Good Sir G 👍🏻😎

Nice job sir👍🔥

by inspection, 3^4 = 81 + 4 = 85

(x-1)(x-2)(x-8)(x+5)+360 = 0 (x^2 -3 x +2)(x^2 -3 x -40) + 360 = 0 (x^2 -3x)^2 -38(x^2 -3 x)+280 = 0 (x^2 -3 x -10)(x^2 -3 x -28) = 0 (x +2)(x -5)(x +4)(x -7) = 0 (x + 4)(x + 2)(x -5)(x -7) = 0 x = -4 Or x = -2 Or x = 5 Or x = 7

another way to eliminate complexity is: x^4 - 2x^2 +1 - 2x^3 + 2x - 0 ( x^2 - 1 )^2 - 2x ( x^2 - 1 ) = 0 ( x^2 - 1 ). ( x^2 - 2x - 1 ) = 0 x^2 - 1 = 0 ==> x = ± 1 or ( x^2 - 2x - 1 ) = 0 x = 1 ± √2

G.K Mathematics Tricks mistake is that he is assuming that e^(i 2k π) = 3 for k= 1,2,3... is true. It is not. 1^(-(ln(3i)/(2π)) = 1 not 3

Komplikovano !!

I watched a vid; when it ended it made several suggestions, I chose this one. On the suggestion-screen was a slightly different problem than you are solving here. I think you should update the suggestion screen to make it match.

(9)^sin^2^x+(9)cos^2^x= {18sin^x^2+18cos^x^2}=36sincosx^4:18^18sincosx^4 3^63^6sincosx^4 1^3^2^1^3^2sincosx^2^2 1^1^3^1sincosx^1^2 3x^2 (sincosx ➖ 3sincosx+2).

2*3ˣ = 2*3²*5 3ˣ = 3²*5 x = 2 + log₃(5) = 2 + log(5)/log(3) ≈ 3.4649735207179...

2ˣ = 7ˣ⁺² x*log₇(2) = x + 2 x = 2/(log₇(2) - 1) = -2/(1 - log(2)/log(7)) ≈ −3.106589511330...

5*x² - 8*x + 3 = 0 , x ≠ 0 (5*x - 3)*(x - 1) = 0 x = 3/5, 1

x⁴ = (x - 2)⁴ Full binomial expansion and factoring: On both sides, drop x⁴ for cubic; 3 solutions. x⁴ = x⁴ - 4*x³*2 + 6*x²*4 - 4*x*8 + 16 0 = -8*(x³ - 3*x² + 4*x - 2) x³ - 3*x² + 4*x - 2 = 0 (x - 1)*(x² -2*x + 2) = 0 x = 1, 1 ± i Factoring difference of squares: x⁴ - (x - 2)⁴ = 0 [x² - (x - 2)²]*[x² + (x - 2)²] = 0 [x - (x - 2)]*[x + (x - 2)]*[2*x² - 4*x + 4] = 0 [2]*[2*x - 2]*[2]*[x² - 2*x + 2] = 0 8*[x - 1]*[(x - 1)² + 1] = 0 x = 1, 1 ± i Roots of 1: ((x - 2)/x)⁴ = 1 = r⁴ , where r = ±1, ±i 1 - 2/x = r x = 2/(1 - r) = 2/0, 2/2, 2/(1 - i), 2/(1 + i) = ∅, 1, 1 + i, 1 - i x = 1, 1 ± i

Let y = 2ˣ . For x ∈ ℝ , y ∈ ℝ and y > 0 . y³ + y² = 12 y*y*(y + 1) = 2*2*(2 + 1) y = 2 -- factor out (y - 2) y³ + y² - 12 = (y - 2)*(y² + 3*y + 6) = 0 y = 2, (-3 ± i*√15)/2 { Discard y ∉ ℝ , y ≤ 0 } 2ˣ = 2 x = 1 2ˣ = (-3 ± i*√15)/2 = r*e^(i*θ) r = √6 , θ = ±arccos(-√6/4) + 2*π*k , k ∈ ℤ x = ln(6)/ln(2)/2 + i*(2*π*k ± arccos(-√6/4))/ln(2)

1ˣ = 3 x*log(1) = log(3) x*0 = log(3) 0 = log(3) { False! } No solution.

x^(1/2 + 1/4 + 1/8 + 1/16) = 4 x^(15/16) = 4 x = 4*¹⁵√4 = 2^(32/15) ≈ 4.3872999187785...

8*(2ˣ)² - 9*2ˣ + 1 = 0 (8*2ˣ - 1)*(2ˣ - 1) = 0 8*2ˣ = 1 , 2ˣ = 1 { log₂(...) } 3 + x = 0 , x = 0 x = -3, 0

This is a fun puzzle I created. Find x, where x! = 3!*6!*22!/11! .

(x - 2)! = 3!*5!*7! = 6*120*7! = 6*15*8*7! = 2*5*9*8*7! = 10*9*8*7! = 10! x = 12

2ˣ + x = 1034 2ˣ = 1034 - x LHS: Power curve from above the asymptote of 0 for x < 0 , climbs to 1 at x = 0 , and goes unbounded for x > 0 . RHS: Line sloping down from (0, 1034) to (1034, 0) . They intersect only once. 2¹⁰ = 1034 - 10 1024 = 1024 x = 10

5ˣ + x = 128 5ˣ = 128 - x LHS: Power curve from above the asymptote of 0 for x < 0 , climbs to 1 at x = 0 , and goes unbounded for x > 0 . RHS: Line sloping down from (0, 128) to (128, 0) . They intersect only once. 5³ = 128 - 3 125 = 125 x = 3

(x - 5)! = 720*7 = 5040 = 7! x - 5 = 7 x = 12

((√5 - 1)/2)⁶ = ( ((√5 - 1)/2)³ )² = ( (5*√5 - 15 + 3*√5 - 1)/8 )² = ( (8*√5 - 16 )/8 )² = (√5 - 2)² = 9 - 4*√5 { √81 - √80 > 0 } ≈ 0.055728090000841...

Nice trick to understand

A nice factorial equation simplification: (x - 2)! = (3!)(5!)(7!); x =? (x - 2)! = (3!)(5!)(7!) = (6)(120)(7!) = (720)(7!) = (10)(9)(8)(7!) = 10! x - 2 = 10; x = 12

1

Good effort

4.2^(2x+1)-9(2^x)+1=0 8(2^x)²-9(2^x)+1=0 2^x=(9±7)/16 --> 2^x=1 x=0 2^x=⅛ --> x=-3 Therefore x={-3,0}

3!5!7!=2.3.2.3.4.5.7! =(2.4)(3.3)(2.5)7! =7!8.9.10 =10! (x-2)!=10! --> x=12

27^x=x⁹ --> (3^x)³=(x³)³ 3^x=x³ --> x=3

Just by looking the problem it is clear that (x-2)!=9! as 8.7.6!=8! and 3!=6. Therefore x-2=9 --> x=11

Posted without seeing the solution. x = 11. You should see immediately that 6! x 7 x 8 = 8!. and 3! = 6.So the 8! and 3! in the numerator cancel out with the denominator. You are left with (x-2)!=9!, so x-2 = 9 and thus x=11.

X=2

ezpz and i am drunk as shi

1 raised to any power (real or complex) is still = 1. It's impossible to find even a complex solution for this problem. No solution.

its 3.