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В уме, без записи - 2😅❤

No era necesario aplicar logaritmo. Aplicando propiedades de las ecuaciones exponenciales ==> n=1

X=0.....May be 2^x=1 2^x=2^0 X=0

the lhs is monotonic increasing, so there is only one real solution. Find the greatest integer value of x that gives 3^x< 248 3^5=243, so subtracting that from each side we have x=5, this confirms x=5 as the only solution. 3^5+5=248.

(x-2)!=3!*5!*7! (x-2)!=3!*3!*4*5*3!*4*5*6*7 (x-2)!=6^3*4^2*5^2*6*8 (x-2)!=216*16*25*6*7 (x-2)!=3628800 (x-2)!=9! x-2=9 x=11

1

Nice logic

Creative logic❤

Well done👍

slabo widac

Х=1...

It is not too difficult to guess it is 10! And hence X=12 as there are two 5 and no 11 with 3!5!7!

bài giảng của ban. rất hay

nice, bro

X=12 Explanation (X-2)!=3!×5!×7! =(3!×5!)×7! =(1×2×3×1×2×3×4×5)×7! ={(2×4)×(3×3)×(2×5)×1}×7! =7!×8×9×10 =10! So, (X-2)!=10! So, X-2=10 X=10+2=12 Hence x=12

^=read as to the power *=read as square root 2^x+4^x+8^x=14 (2^x)+(2^2x)+(2^3x)=14 Let 2^x=R So, R+R^2+R^3=14 Add 1 to the equation R^3+R^2 +R +1=14+1 (R^2(R+1)+1(R+1)=15 (R^2+1)(R+1)=15 We know 15=1×15 =15×1 =3×5 =5×3 and many more So, Case 1 (R^2+1)(R+1)=1×15 So, R^2+1=1, orR+1=15 R^2=1-1 or R=15-1 R^2=0 or R=14 R=0 or R=14 So, 2^x=0, or 2^x=14 If 2^x=14 X=log14 of base 2 Case 2 will produce same result (R^2+1)(R+1)=3×5 R^2+1=3 or2 R+1=5 R^2=3-1 or R=5-1 R^2=2 or R=4 R=*2 or R=4 If R=*2 2^x=2^(1/2) X=1/2 If R=4 2^x=2^2 X=2 But x does not satisfies the equation Case 3 (R^2+1)(R+1)=5×3 So, R^2+1=5 or R+1=3 R^2=5-1 or R=3-1 R^2=4 or R=2 R=2 or R=2 So, 2^x=2 2^x=2^1 X=1 X =1 in both situations Hence X=1 is much accepted...

x=3

x=2

very nice

3. This one's easy. Solved in 2 sec.

3!5!7! = 2 3 2 3 4 5 2 3 4 5 6 7 = 2 3 4 5 6 7 2 3 4 5 2 3 = 2 3 4 5 6 7 8 9 10 = (12 -2)! x = 12

X^X^2=16=2^4=2^2^2 X=2

(X-2)!=3!×5!×7!=6×5×4×3×2×1×7!=7!×8×9×10=10! X-2=10 X=12

This isn't much of a trick as it is... just the correct way to solve it. Prime Newtons has a video explaining the Lambert W function with a very similar equation, and the method used is identical to yours. Then, once you look at the initial equation and spot that x=2 must be a solution, it's obvious that there must be some way of manipulating everything inside W to remove it entirely and end up with a nice, neat solution. Once you know the method, the rest is straightforward enough.

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