4^x - 2^x+1 = 3 How do you solve this?

2^(2x) - 2^(x + 1) - 3 = 0
2^(2x) - 2 (2^x) - 3 = 0
let y = 2^x
y ² - 2 y - 3 = 0
( y - 3 ) ( y + 1 ) = 0
y = 3 , y = - 1

Accepting y = 3 :-
2^x = 3
x ln 2 = ln 3
x = ln 3 / ln 2
x = 1.58 (to 2 dec. places)

math_kp is right

4^x = (2^2)^x = (2^x)^2 = t^2 for t=2^x

and 2^(x+1) = 2*2^x = 2t

that's where t^2 - 2t - 3 = 0 came from, so you should get x=log(3)/log(2) -- which is log 3 base 2, or ln(3)/ln(2).

Assume t =2^x
t^2-2t-3 = 0
(t-3)(t+1) = 0
t = 3, 2^x = 3, Take log on both sides. x = log(3)/log 2
t = -1, 2^x should be alwasy > 0. No solution for x.

Let t =2^x
t^2-2t-3 = 0
(t-3)(t+1) = 0
t = 3, 2^x = 3, Take log on both sides. x = log(3)/log 2
t = -1, it is impossible for 2^x <0. So, no solution for real x.

group like terms ans solve for x
4^x - 2^x=3-1 remember the sign changes when it crosses the equals sign
now 4^x- 2^x= 2^x
and 3-1=2 so the new equation is
2^x=2 solve for x by dividing by both sides by 2
x=1

I think it's supposed to be 4x-2x+1=3 so it is 2x+1=3
2x=2
therefore, x=1

2^x = t
t^2-2t-3 = 0
(t-3)(t+1) = 0
t = 3 means x = log(3)/log 2
t = -1 means no solution real x

4^x-2^x+1=3
4^x-2^x-2=0
let t=2^x
=> t^2-t-2=0
=> t=2 &t= -1

It's obvious: x = 1.

4 - 2 + 1 = 3

OK, if the "x + 1" is in the exponent of the second term, then it should be written as 2^(x + 1) to avoid confusion. Hint "write as a quadratic" probably also means you left off the "square" and it should be 4 to the x-square power. Otherwise you have a nonsense equation because 4^x IS 2^(x+1)
and you are trying to "solve" x - x = 3, which is false every time.

4^x - 2^x+1 = 3
4^x -2^x -2 =0
let t = 2^x
t^2 -2t -2 =0
solve this for t
and find 2^x =t
xln(2) =ln(t)
x = ln(t)/ln(2)