please,answer! thanks

Penyelesaian:

a. [tex] \sqrt{( \frac{1}{(9})^x+3 } = \sqrt[3]{27} [/tex]
= [tex] \sqrt{( \frac{1}{(9})^x+3 } = 3[/tex]
= [tex] (\frac{1}{9})^x+3 = 9 [/tex]  (kedua ruas di kuadratkan)
= (9^-1)^x+3 = 9^1
= 9^-x-3 = 9^1
-x - 3 = 1
x = -4

b. [tex] \frac{ \sqrt{16^x+2} }{32} = 1[/tex]
= [tex] \sqrt{16^x+2} = 32[/tex]
= 16^x+2 = 2^10  (kedua ruas di kuadratkan)
= (2^4)^x+2 = 2^10
= 2^4x+8 = 2^10
4x + 8 = 10
4x = 2
x = 1/2