Question

Buna, ma poate lamuri cineva cu aceasta ecuatie 3 la puterea x+1= 9 la puterea radical din x.

Answer 1

   
[tex]\displaystyle \bf\\ 3^{x+1} = 9^{ \sqrt{x} }\\\\ 3^{x+1} = \Big(3^2\Big)^{ \sqrt{x} }\\\\ 3^{x+1} = 3^{2 \sqrt{x} }\\\\ x+1 = 2\sqrt{x} ~~~\Big| ~Ridicam~ la ~puterea ~ a~ doua.\\\\ (x+1)^2=(2\sqrt{x})^2\\\\ x^2+2x+1 = 4x\\\\ x^2+2x-4x+1 = 0\\\\ x^2 -2x+1 = 0\\\\ (x-1)^2 = 0\\\\ x-1 = 0\\\\ \boxed{\bf x = 1 } ~~~\text{(Radacina dubla)} [/tex]