A nice algebra problem to solve| Math Olympiad

[@mathquizzes1]

good

[@mr.x2870]

3:42 wrong. 2^0=1. Same with 3. So LHS =2 which is equal to RHS.

[@eliagerber4103]

X≠0 because 6^x-3^x≠0 from the beginning equation.

[@IsaacVázquez-l7m]

Siempre siempre siempre debes realizar tu verifica en la ecuación original, no en una ecuación derivada de la original.

[@lourdesgracia6767]

Por simple revisión visual x=1

[@GeorgetaBobar]

X=1

[@SidneiMV]

2ˣ(4ˣ - 1)/[3ˣ(2ˣ - 1)] = 2 x ≠ 0 ∈ ℤ 2ˣ(2ˣ + 1)/3ˣ = 2 2ˣ⁻¹(2ˣ + 1) = 3ˣ [ I ] for x > 0 2ˣ⁻¹ = 1 x - 1 = 0 *x = 1* 2ˣ + 1 = 3ˣ 2 + 1 = 3 (true) [ II ] for x < 0 x = -u => u > 0 2⁻ᵘ⁻¹(2⁻ᵘ + 1) = 3⁻ᵘ 2⁻ᵘ⁻¹(2⁻ᵘ + 1) = 3⁻ᵘ 2⁻ᵘ⁻¹(2ᵘ + 1)/2ᵘ = 3⁻ᵘ 2⁻²ᵘ⁻¹(2ᵘ + 1) = 3⁻ᵘ 3ᵘ(2ᵘ + 1) = 2¹2²ᵘ 2ᵘ + 1 must be even => u = 0 but u > 0 => no solution for x < 0

[@mks-learning-easy]

Appreciate your knowledge 😊. Thank you ❤️❤️

[@SidneiMV]

@@mks-learning-easy thanks. but I didn't analyze for x < 0. I will do it.

[@varshithpudota2578]

So x=1

[@Xjxtimvbhf-ujnjdmczrdjqyt]

Охуенно ты сокращаешь - где О.О.?

[@elenaokon2736]

So, solution assumes that x is an integer ?

[@N_Kumar26]

2^×(2^× + 1)=2(3^×) ×=1 L.H.S=R.H.S

[@КатяРыбакова-ш2д]

x=1