Bonjour,
[tex]4(ln(x))^2-ln(x^2)-7=0\\
4(ln(x))^2-2ln(x)-7 = 0\\\\
En \ posant \ X=ln(x)\\\\
4X^2-2X-7=0\\
\Delta = (-2)^2-4\times4\times (-7)=116\\\\
X_1= \frac{2- \sqrt{116} }{2\times 4}= \frac{1- \sqrt{29} }{4} \\\\
X_1= \frac{2+\sqrt{116} }{2\times 4}= \frac{1+\sqrt{29} }{4} \\\\
D'ou : \\\\
ln(x_1) = \frac{1- \sqrt{29} }{4} \Longrightarrow \boxed{x_1=e^{\frac{1- \sqrt{29} }{4}}}\\\\
ln(x_2) = \frac{1+\sqrt{29} }{4} \Longrightarrow \boxed{x_2 = e^{\frac{1+ \sqrt{29} }{4}}}[/tex]