Solve for x (2^2x+4 - 6.2^2x+1) / 2^x+1 = 16
I'm going to put some brackets in this expression also. To give
(2^(2x+4) - 6*2^(2x+1)) / 2^(x+1) = 16
Using 2^(2x+4) = 2^(2x)*2^4, etc. the expression becomes
(2^(2x)*2^4 - 6*2^(2x)*2^1) / (2^(x)*2^1) = 16
(16*2^(2x) - 12*2^(2x)) / (2*2^(x)) = 16
(16*2^(2x) - 12*2^(2x)) = 16*(2*2^(x))
4*2^(2x) = 32*2^(x)
2^(2x) = 8*2^(x)
Let u = 2^x, then
u^2 = 8u
u^2 - 8u = 0
u(u - 8) = 0
u = 0, u = 8
Therefore, 2^x = 0, 2^x = 8
i.e. x = -infinity, x = 3
x = -infinity is a solution of the quadratic eqn, but not of the original expression, so we ignore it.
The solution is: x = 3