Resposta:

3^(x⁴) =81ˣ *3

3^(x⁴) =(3⁴)ˣ *3

3^(x⁴) =3⁴ˣ * 3¹

3^(x⁴) =3⁴ˣ ⁺¹

x⁴ =4x+1

x⁴ -4x-1=0

a=1 , b=0  , c=0  , d=-4  , e = -1

x₁=1.66325193877147

x₂=−0.24903837639837

x₃=−0.70710678118655+1.3835510696657i

x₄=−0.70710678118655−1.3835510696657i

Verified answer

[tex]\displaystyle \sf \frac{3^{x^4}}{81^x} = 3 \\\\\ \frac{3^{x^4}}{(3^4)^x}= 3 \\\\\ 3^{x^4-4x} =3 \\\\\ x^4-4x -1=0 \\\\\ \text{sabemos que}: \\\\ (x^2+1)^2=x^4+2x^2+1 \\\\\ x^4 = (x^2+1)^2-2x^2-1 \\\\\ \text{substituindo na equa\c c\~ao} :\\\\ (x^2+1)^2-2x^2-1-4x-1 = 0 \\\\\ (x^2+1)^2-2x^2-4x-2 = 0 \\\\\ (x^2+1)^2-2(x^2+2x+1)=0 \\\\\ (x^2+1)^2-2(x+1)^2= 0 \\\\\ (x^2+1)^2=2(x+1)^2 \\\\\ \sqrt{(x^2+1)^2} =\pm\sqrt{2(x+1)^2} \\\\ x^2+1 = \pm\sqrt{2}(x+1)[/tex]

[tex]\displaystyle \sf 1\º caso : \\\\\ x^2+1=\sqrt{2}(x+1) \\\\ x^2+1=x\sqrt{2}+\sqrt{2} \\\\\ x^2-x\sqrt{2}+1-\sqrt{2}=0\\\\\ x = \frac{-(-\sqrt{2})\pm\sqrt{(-\sqrt{2})^2-4\cdot (1-\sqrt{2})}}{2} \\\\\\ x=\frac{\sqrt{2}\pm\sqrt{2-4+4\sqrt{2}}}{2} \\\\\ \boxed{\sf x = \frac{\sqrt{2}\pm\sqrt{4\sqrt{2}-2}}{2} \ }[/tex]

[tex]\displaystyle \sf 2\º caso : \\\\\ x^2+1=-\sqrt{2}(x+1) \\\\ x^2+1=-x\sqrt{2}-\sqrt{2} \\\\\ x^2+x\sqrt{2}+1+\sqrt{2}=0\\\\\ x = \frac{-\sqrt{2}\pm\sqrt{(\sqrt{2})^2-4\cdot (1+\sqrt{2})}}{2} \\\\\\ x=\frac{\sqrt{2}\pm\sqrt{2-4-4\sqrt{2}}}{2} \\\\\ x = \frac{\sqrt{2}\pm\sqrt{-4\sqrt{2}-2}}{2} \\\\\\ \text{note que} : \\\\\ -4\sqrt{2}-2 < 0 \to \sqrt{-4\sqrt{2}-2} \notin \mathbb{R} \to \text{N\~ao conv\'em }[/tex]

Portanto as soluções são :

[tex]\displaystyle \boxed{\sf \sf x = \frac{\sqrt{2}+\sqrt{4\sqrt{2}-2}}{2} \ ; \ x = \frac{\sqrt{2}-\sqrt{4\sqrt{2}-2}}{2} \ }\checkmark[/tex]

comentário :

caso a questão quisesse todas as soluções, bastaria resolver o 2º caso p/ soluções nos complexos