indices question?

Question

2^(2x+1)-3(2^x)+1 = 0

solve for x using substitution method of y=a^x

could somebody please help me with this question?

Answer 1

Expand the first exponential term:

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

So your equation becomes

2*(2^x)^2 - 3*(2^x) + 1 = 0

Now substitute y=2^x to get

2*y^2 - 3*y + 1 = 0

Factor this

(2y-1)*(y-1) = 0

y = 0.5, y=1

Now put y=2^x into these solutions
2^x = 0.5 and 2^x = 1

By inspection, x = -1 and x = 0

2^-1 = 1/2 = 0.5;
2^0 = 1

Answer 2

2^(2x).2 - 3.2^(x) + 1 = 0

Let y = 2^(x)

2y² - 3y + 1 = 0

(2y - 1).(y - 1) = 0

y = 1/2, y = 1

2^(x) = 1/2 and 2^(x) = 1

x= -1, x = 0