2^{2x}-33\cdot 2^x+32=0 this...

The problem

2^{2x}-33\cdot 2^x+32=0

this equation doesnt give me a solution?

the answer must be 0 or 5

Replace 2^x with y, that is, let

Now, we need to solve the following equation:

y^2 + 33y + 32 = 0

........

........

y1 = 1

y2 = 32

For y1 = 1 we have:

2^x = 1

........

x = 0

For y2 = 32 we have:

2^x = 32

.........

x = 5

The answers indeed are 0 and 5.