Assuming that we're not getting into complex numbers here, there's really no need to do any computation, rearrangement, subsititution, etc. `2^x` is positive for all `x` , `4^x` is positive for all `x` , and `4` is positive, so the expression `2^x+4^x+4` is always positive, never `0.`

The equation `2^x + 4^x + 4 = 0` is equaivalent to `2^x + (2^2)^x + 4 = 0`

=> `2^x + (2^x)^2 + 4 = 0`

If `2^x = y`

=> `y^2 + y + 4 = 0`

The roots of this quadratic equation are `(-1+-sqrt(1 - 16))/2` = `-1/2 +- i*(sqrt 15)/2`

As `2^x = y` , the value of y is real for any value of x. As this is not the case here, the given equation does not have a solution.

**The equation `2^x + 4^x + 4 = 0` does not have a solution as a positive real number raised to any power gives a real result.**

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