Bonjour belgacem ;
On a : y = 5x ;
donc : x^y = y^x <=> x^(5x) = (5x)^x <=> Ln( x^(5x)) = Ln((5x)^x)
<=> 5x Ln(x) = x Ln(5x) <=> 5 Ln(x) = Ln(5x) = Ln(5) + Ln(x)
<=> 4 Ln(x) = Ln(5) <=> Ln(x) = (Ln(5))/4
<=> x = exp((Ln(5))/4) = (exp(Ln(5)))^(1/4) = (5)^(1/4) .
On a donc : x² = ((5)^(1/4))² = (5)^(1/2) ;
donc : xy = 5x² = 5 * (5)^(1/2) = (5)^(3/2) .