Germany | Math Olympiad Problem | Fully explained

[@AimanIssamQatar]

good

[@SidneiMV]

2²ˣ = (2x)³²
(2x)ln2 = 32ln(2x)
(1/2x)ln(2x) = (1/32)ln2
[1/(2x)]ln[1/(2x)] = (1/32)ln(1/2)
[1/(2x)]ln[1/(2x)] = (1/2⁵)(1/2³)8ln(1/2)
[1/(2x)]ln[1/(2x)] = (1/2⁸)ln(1/2⁸)
1/(2x) = 1/2⁸ => *x = 2⁷ = 128*

[@MasterTMathsClass]

Nice

[@cassioburgos1159]

Can you explain your resolution from the 3rd. line onwards? I really dont understand. If possible, thanks.

[@SidneiMV]

@@cassioburgos1159 step by step
(1/2x)ln(2x) = (1/32)ln2
(1/2x)(-1)ln(2x) = (1/32)(-1)ln2
[1/(2x)]ln[1/(2x)] = (1/32)ln(1/2)
[1/(2x)]ln[1/(2x)] = (1/2⁵)ln(1/2)
(1/2⁵)ln(1/2) = (1/2⁵)(1/a)(a)ln(1/2)
= 1/(a2⁵)ln(1/2ᵃ)
let a2⁵ = 2ᵃ
2ᵃ⁻⁵ = a => a = 8 (by inspection)
[1/(2x)]ln[1/(2x)] = (1/2⁵)(1/8)(8)ln(1/2)
[1/(2x)]ln[1/(2x)] = (1/2⁵)(1/2³)ln(1/2⁸)
[1/(2x)]ln[1/(2x)] = (1/2⁸)ln(1/2⁸)
applying Lambert W function both sides
W{[1/(2x)]ln[1/(2x)]} = W[(1/2⁸)ln(1/2⁸)]
ln[1/(2x)] = ln(1/2⁸)
1/2x = 1/2⁸ => 2x = 2⁸ => *x = 2⁷ = 128*

[@СергейИванчиков-р8д]

Good

[@MasterTMathsClass]

Thanks