3 ^(2x +1) - 28(3^x) +9 =0 ,find x?

Answer 1

3^x =u
3^(2x+1)-28(3^x)+9=0
3^2x=u^2
3^1=3
You add powers when you multiply
3^2x*3^1=3^(2x+1)
therefore we get to

3u^2-28u+9=0

u=9 , u=1/3

3^x=9
x=log9/log3
x=2

3^x=1/3
x=lod1/3.log3

x=-1

Answer 2

3^(2x + 1) --28(3^x) + 9 = 0
=> 3(3^x)^2 --28(3^x) + 9 = 0
=> (y --9)(3y --1) = 0 where y = 3^x
giving 3^x = 9 = 3^2 and 3^(x +1)= 1 = 3^0
whence x = 2, --1