xˣ=(5/6)²
posso scrivere
ln(xˣ)=ln(5/6)²
xln(x)=ln(5/6)²
x₁=exp(W₀(xln(x))=exp(W₀(-2ln(6/5)))≈0.417743793241348456...
per 1/e≤x
x₂=exp(W₋₁(xln(x)))=exp(W₋₁(-2ln(6/5)))≈0.320173486649844631...
per
0<x≤1/e
specifico e preciso:
xˣ=(5/6)²
posso scrivere
ln(xˣ)=ln(5/6)²
xln(x)=ln(5/6)²
x₁=exp(W₀(xln(x))=exp(W₀(-2ln(6/5)))≈0.417743793241348456...
per 1/e≤x
x₂=exp(W₋₁(xln(x)))=exp(W₋₁(-2ln(6/5)))≈0.320173486649844631...
per
0<x≤1/e