9^x + 4*3^x - 3 = 0. Can anyone help me solve this please?

Answer 1

3^2x +4*3^x--3=0

let 3^x=a

a^2+4a-3=0

CORRECTION
a^2-4a-3=0

(a-3)(a-1)=0

3^x=3
3^x=1

so

x=1
x=0

Not correct unless +4 is changed to -4
I don't know how to solve it otherwise.

Answer 2

The point is that 9^x = 3^(2x) = (3^x)^2.

Thus, the problem is a quadratic equation in 3^x.

Let T = 3^x. Then T^2 +4T - 3 = 0. You'll have to use the quadratic formula to get two roots of that.

Once you know T (two choices -- both roots are real), then x = ln (base 3) T.

Answer 3

9^x+4*3^x-3=0
taking logarithm
you can solve
x log 9+x log 12-log 3=0
try it reply me

Answer 4

x= ln3/(ln9+ln12)