What are the solutions of the equation 2*16^x = 4^x + 1 ?

We notice that 16=4^2!

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

2*(4^2)^x - 4^x -
1=0

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

2*t^2 - t - 1=0

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

t1=[1+sqrt
(9)]/4

t1=(1+3)/4

t1=1

t2=[1-sqrt
(1+4*2)]/4

t2=(1-3)/4

t2=-1/2

We
didn't find the values of x,
yet!

4^x=1

4^x=4^0

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

x
= 0

4^x=-1/2

The exponential
4^x is always positive, for any value of x, so, we'll reject the second
solution.

The equation has just one solution.
The only solution is x= 0.