What are the roots of the equation 16^x-3*4^x+2=0?

We'll write 16 = 4^2.

We'll
re-write the equation in this manner:

(4^2)^x - 3*4^x + 2 =
0

We'll substitute 4^x by the variable,
t.

t^2 - 3t + 2 = 0

We'll
apply quadratic formula for finding t:

t1 = [3+sqrt
(9-4*2)]/2*1

t1 = [3+sqrt
(1)]/2

t1 =
(3+1)/2

t1 =
2

t2 =
(3-1)/2

t2 =
1

We'll find the values of x,
now.

4^x =
t1

4^x=2

We'll write 4^x =
2^2x

2^2x = 2^1

Since the
bases are matching, we'll apply one to one property:

2x =
1

x =
1/2

4^x =
t2

4^x = 1

4^x =
4^0

x =
0

The solutions of the
equation are {0 ; 1/2}.