diketahui persamaan
[tex] \sqrt{ {3}^{2x + 1} } = {9}^{x - 2} [/tex]
tentukan nilai X![]()
[tex] \sqrt{ {3}^{2x + 1} } = {9}^{x - 2} [/tex]
tentukan nilai X
Penjelasan dengan langkah-langkah:
[tex] \sqrt{ {3}^{2x + 1} } = {9}^{x - 2} [/tex]
[tex] {3}^{ \frac{2x + 1}{2} } = {3}^{2.(x - 2)} [/tex]
[tex] \frac{2x + 1}{2} = 2.(x - 2)[/tex]
[tex]2x + 1 = 4.(x - 2)[/tex]
[tex]2x + 1 = 4x - 8[/tex]
[tex]2x - 4x = - 8 - 1[/tex]
[tex] - 2x = - 9[/tex]
[tex]x = \frac{9}{2} [/tex]
Penyelesaian:
√(3^(2x+1))=9^x-2
3^½(2x+1)=3^2(x-2)
x+½=2(x-2)
x+½=2x-4
4½=x
[tex] \boxed{ \colorbox{black}{ \sf{ \color{lightgreen}{ answered\:by\:Duone}}}} [/tex]
[answer.2.content]