We figured that this can be changed to $2^{2x} - 2^x \cdot 2 = 3$, but couldn't solve from there. Perhaps we are not on the right path?
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2from the two answers so far, you seem to be on the right track. – robjohn Mar 15 '13 at 21:47
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You are on the right path.
Substitute $y=2^x$, you get $y^2-2y-3=0$. Try to solve that and change it back to $x$.
NECing
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$4^x - 2^{x + 1} = 3$ $$(2^{x})^2 - 2\cdot2^{x}-3 =0,\qquad 2^x=t$$ $$t^2-2t-3=0$$ $$t_{1,2}=3,-1$$ $$2^x=3\Rightarrow x=\log_2 3$$
Adi Dani
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2If $x$ is allowed to be non-real, for the case $2^x = -1$ you also get the set of solutions $x=\frac{i\pi(2n+1)}{\ln(2)}$ for $n\in\mathbb{Z}$. – Peter Olson Mar 16 '13 at 01:50