A Very Nice Math Olympiad Problem | Solve for x | You Need To Know This Trick | Algebra

In this video, I'll be showing you step by step on how to solve this Olympiad Maths Exponential problem using a simple trick.
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[@aperson2020]

Nice.

[@SpencersAcademy]

Thank you 😊

[@tommyliu7020]

This feels like an incredibly slow method. Once we have it in the form of x^6 = 2^6, we know the answers are the 6th roots of unity, scaled by the magnitude of 2. So basically, the 6 answers are: 2e^(i*0) = 2 2e^(i*pi/3) 2e^(2i*pi/3) 2e^(i*pi) = -2 2e^(4i*pi/3) 2e^(5i*pi/3) You can figure this out in like 15 seconds.

[@ربهامابراهيم-ح1ن]

رائعه

[@SpencersAcademy]

Thank you 🙏

[@niranjanchakraborty1139]

Ans=1. x^3/8=8/x^3=1

[@williamBryan-k2e]

can someone explain to me why you cannot just do this x^5 = 2^2 = now obviously x=2 ( and actually +/- 2 - as we have an even exponent. this is not unlike when you say 2^x = 2^y - so use the rule of exponents to say x=y.

[@BluesChoker01]

The rule of exponents requires terms have the same base, as you know. Here, we have two real solutions and four complex solutions, so I don't know that this is the exact approach that would help you solve them all. Now, you could take the 6th root of each side, and generate X = +-2 for the reals (as the beginning factors have an even number, or six intercepts in the Real and Complex planes (X^6 = 2^6). Better still, if you have mastered the nth root of unity (1) technique, you could use that advanced technique to solve all the roots. Look online or in a textbook to read about it. Now using exponents to produce a difference of squares--which then can produce a sum and difference of cubes--along with the linear terms for the real solutions, can be done fast with practice. Unless you're really quick at expansions to factors and the reverse, it's propably worth doing. And again, it's a shortcut to producing one equation, fully expanded, to solve for all 6 intercepts. Check my post for an example. Only needed figure one complex solution, as I got the pattern. Anyway, have fun, and yes, it's always good to look for alternative approaches. Even a dead end will learn you something. 👍

[@SpencersAcademy]

Excellent 👌

[@vincentrobinette1507]

X=2 2/2=1 1X1X1=1 1X1=1

[@phillipmcduffie9353]

2 ?

[@श्रीस्वामीसमर्थ-थ7घ]

x = 2 And when we simplify equation by putting x =2 then answer is 1

[@key_board_x]

(x/2).(x/2).(x/2) = (2/x).(2/x).(2/x) → where: x ≠ 0 (x/2)³ = (2/x)³ (x/2)³ - (2/x)³ = 0 → recall: a³ - b³ = (a - b).(a² + ab + b²) [(x/2) - (2/x)].[(x/2)² + (x/2).(2/x) + (2/x)²] = 0 [(x²/2x) - (4/2x)].[(x²/4) + 1 + (4/x²)] = 0 [(x²/2x) - (4/2x)].[(x⁴/4x²) + (4x²/4x²) + (16/4x²)] = 0 [(x² - 4)/2x)].[(x⁴ + 4x² + 16)/4x²] = 0 (x² - 4).(x⁴ + 4x² + 16)/8x³ = 0 (x² - 4).(x⁴ + 4x² + 16)/x³ = 0 → recall: x ≠ 0 (x² - 4).(x⁴ + 4x² + 16) = 0 (x + 2)(x - 2).[x⁴ + 4x² + 16] = 0 (x + 2)(x - 2).[(x² + 2)² - 4 + 16] = 0 (x + 2)(x - 2).[(x² + 2)² + 12] = 0 First case: (x + 2) = 0 → x = - 2 Second case: (x - 2) = 0 → x = 2 Third case: [(x² + 2)² + 12] = 0 (x² + 2)² + 12 = 0 (x² + 2)² = - 12 (x² + 2)² = 12i² x² + 2 = ± i√12 x² + 2 = ± 2i√3 x² = - 2 ± 2i√3 First possibility: x² = - 2 + 2i√3 x² = - 2 + 2i√3 x² = 1 - 3 + 2i√3 x² = 1 + 2i√3 - 3 x² = 1 + 2i√3 + 3i² x² = (1)² + 2i√3 + (i√3)² x² = (1 + i√3)² → x = ± (1 + i√3) Second possibility: x² = - 2 - 2i√3 x² = - 2 - 2i√3 x² = 1 - 3 - 2i√3 x² = 1 - 2i√3 - 3 x² = 1 - 2i√3 + 3i² x² = (1)² - 2i√3 + (i√3)² x² = (1 - i√3)² → x = ± (1 - i√3)

[@SpencersAcademy]

Your maths skills are really exceptional. Great job 👏