How many solutions has the equation 4^x-2^x=56?

We'll create a common base, so we'll write 4^x =
2^2x.

We'll re-write th
equation:

2^2x - 2^x - 56 =
0

We'll replace 2^x by t:

t^2
- t - 56 = 0

We'll apply quadratic
formula:

t1 = [1 +
sqrt(1+224)]/2

t1 = (1 +
sqrt225)/2

t1 = (1+15)/2

t1 =
8

t2 = (1-15)/2

t2 =
-7

We'll put 2^x = t1 => 2^x =
8

We'll create matching
bases:

2^x = 2^3

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

x =
3

We'll put 2^x = t2 => 2^x = -7
impossible.

The equation will have only one
solution and this one is x = 3.