Can you solve this? | Math Puzzle

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In this video, we'll show you How to Solve Math Olympiad Question A Nice Math Puzzle in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
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[@gopagon2114]

May be in first row 1,15 and second row 2,-3

[@فیروزاهنگری]

ab=15 , a+c=3 , c-d=5 , b+d=12 {c-d=5 , b+d=12 then c+b=17,b=17-c a+c=3 , a=3-c {a=3-c , b=17-c , ab=15 then (c-c)(17-c)=15 , c^2-20c+36=0 (c-18)(c-2)=0 c=18 , b=17-c=17-18 , b=-1 c=2 , b=17-2 , b=15 {ab=15 , b=-1 , a=-15 {ab=15 , b=15. ,. a=1 {c-d=5 ,. c=2 , d=-3 {c-d=5. ,. c=18. , d=13 {a=1 , b=15 , c=2 , d=-3 {a=-15 , b=-1 , c=18 , d=13

[@key_board_x]

(1): ab = 15 (2): a + c = 3 (3): c - d = 5 (4): b + d = 12 From (3) + (4) (c - d) + (b + d) = 5 + 12 c - d + b + d = 17 b + c = 17 → from (1): ab = 15 → b = 15/a (15/a) + c = 17 → from (2): a + c = 3 → c = 3 - a (15/a) + 3 - a = 17 (15/a) - a = 14 15 - a² = 14a a² + 14a - 15 = 0 Δ = 14² - (4 * - 15) = 196 + 60 = 256 = 16² a = (- 14 ± 16)/2 a = - 7 ± 8 a = 1 or a = - 15 First case: a = 1 Recall: b = 15/a → b = 15 Recall: c = 3 - a → c = 2 From (4): b + d = 12 → d = 12 - b → d = - 3 Second case: a = - 15 Recall: b = 15/a → b = - 1 Recall: c = 3 - a → c = 18 From (4): b + d = 12 → d = 12 - b → d = 13

[@SALogics]

Very nice! ❤

[@SALogics]

Can you please solve this equation 8^(x-1)=x-1/6

[@key_board_x]

8^(x - 1) = x - (1/6) 8^(x) * 8^(- 1) = x - (1/6) 8^(x) / 8 = x - (1/6) 8^(x) = 8x - (8/6) 8^(x) = 8x - (4/3) [8x - (4/3)] / 8^(x) = 1 [8x - (4/3)] * 8^(- x) = 1 [(4/3) - 8x] * 8^(- x) = - 1 8.[(4/24) - x] * 8^(- x) = - 1 8.[(1/6) - x] * 8^(- x) = - 1 [(1/6) - x] * 8^(- x) = - 1/8 [(1/6) - x] * 8^(- x) * 8^(1/6) = - (1/8) * 8^(1/6) [(1/6) - x] * 8^[- x + (1/6)] = - 8^(- 1) * 8^(1/6) [(1/6) - x] * 8^[(1/6) - x] = - 8^[- 1 + (1/6)] [(1/6) - x] * 8^[(1/6) - x] = - 8^(- 5/6) [(1/6) - x] * 8^[(1/6) - x] = - [2^(3)]^(- 5/6) [(1/6) - x] * 8^[(1/6) - x] = - 2^(- 5/2) → you can write 8 as e^Ln(8) [(1/6) - x] * { e^Ln(8) }^[(1/6) - x] = - 2^(- 5/2) [(1/6) - x] * e^{ Ln(8) * [(1/6) - x] } = - 2^(- 5/2) [(1/6) - x] * e^{ [(1/6) - x].Ln(8) } = - 2^(- 5/2) Ln(8) * [(1/6) - x] * e^{ [(1/6) - x].Ln(8) } = - [2^(- 5/2)] * Ln(8) → Lambert W function: ω [β.e^(β)] = β Ln(8) * [(1/6) - x] = ω{- [2^(- 5/2)] * Ln(8)} [(1/6) - x] = ω{- [2^(- 5/2)] * Ln(8)} / Ln(8) - x = [ω{- [2^(- 5/2)] * Ln(8)} / Ln(8)] - (1/6) x = (1/6) - [ω{- [2^(- 5/2)] * Ln(8)} / Ln(8)] To go further = 2^(- 5/2) = 2^[(- 4 - 1)/2] = 2^[(- 4/2) - (1/2)] = 2^[- 2 - (1/2)] = 2^(- 2) * 2^(- 1/2) = [1/2^(2)] * [1/2^(1/2)] = (1/4) * (1/√2) = 1/(4√2) = (√2)/8 x = (1/6) - [ω{- (√2)/8 * Ln(8)} / Ln(8)] I hope this help you….

[@reynaldoluna2139]

By trial and error first box is -15 second box is -1 first mu ltiply by -1 equal +15 third box is +13 and fourthbox is (+13) equal 5. First box multiply by second box that is -15 ×- 1=15 Firstbox +third box that is -15 +(+18) =3 Second box +fourth box that is -1 +(+13) =12