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8^(sqrt(x)) = 4^(sqrt(x)) + 1 find x
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\[8^{\sqrt{x}}=4^{\sqrt{x}}+1\] Let's find \[y=2^\sqrt{x}\] first \[y^3=y^2+1 \rightarrow y^3-y^2-1=0\]
Let u = 2^sqrt(x). Then notice that u^2 = 4^sqrt(x) and u^3 = 8^sqrt(x) So u^3 = u^2 + 1 .... u^3 - u^2 - 1 = 0
The unfortunate thing now is that this equation looks simple but doesn't have a particularly nice solution.
u^2 = 2^(2(sqrt(x))
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