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2^(-x^2)=4
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Solve for x: \[2^{-x ^{2}}=4\]
\[\huge 2^{-x^2} = 4\] \[\huge x^2 = -\frac{ \ln 4 }{ \ln 2 }\] after that square both sides.
Yep, you get -2. So you have complex answers. I've been looking at this over and over and they are technically extraneous solutions because you cannot insert those into a logarithm and have it give you an answer.
\[2^{-x ^{2}}=4 or \frac{ 1 }{2^{x ^{2}} }=2^{2},1=2^{x ^{2}}*2^{2}\] \[2^{x ^{2}+2}=1=2^{0},x ^{2}+2=0,x ^{2}=-2=2\iota ^{2},x=\pm \sqrt{2}\iota \]
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