A Nice Exponential System With Logs

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[SidneiMV]

logx = u => x = 10ᵘ
logy = v => y = 10ᵛ

10ᵘᵛ + 10ᵘᵛ = 2 =>10ᵘᵛ = 1 => uv = 0
u = 0 ∨ v = 0 => x = 1 ∨ y = 1

x = 1
1 + y^logy = 11 => y^logy = 10
(logy)² = 1 => y = 10 ∨ y = 1/10

y = 1
1 + x^logx = 11 => x^logx = 10
(logx)² = 1 => x = 10 ∨ x = 1/10

(x, y) = {(1, 10); (1, 1/10); (10, 1); (1/10, 1)}

[mikecaetano]

Kinda looks like the infield surrounding a baseball diamond.

[aurochrok]

it’s easy. done in my head from the thumbnail. all one has to realize is that the first equation is equivalent to 2 x 10^(logx x logy) = 2 => logx x logy = 0

[SweetSorrow]

This problem looks crazy. Definitely requires substitution.