Solve for x
1) 2^3-5x=(1÷64)^x-1
2) 4^2x÷(1÷2)^x=32
3) 2^2x+1-5(2^x)+2=0
#1. Make everything powers of 2:
64 = 2^6, so 1/64^x = 2^-6x
2^3 - 5x = 2^-6x - 1
Now that doesn't get easily solved, so I assume the usual carelessness with parentheses, giving us
2^(3-5x) = (1/64)^(x-1) = 2^(-6(x-1))
3-5x = -6x+6
x = 3
do the others in like wise. Come back if you get stuck, but show us how far you got.
Actually, the others are easier.
#2 is just an exercise in simplifying fractions
#3 is just a quadratic equation in 2^x. Try setting u=2^x and you have
2u^2 - 5u + 2 = 0