Answer:
[tex]x = \frac{log\sqrt{-1/6}}{log2}[/tex]
Step-by-step explanation:
Given the expression
[tex]2^{2x}+3-7(2^{2x}+1)+3=0[/tex]
Let [tex]P=2^x[/tex]
Substituting into the expression, we will have:
[tex]P^2+3-7(P^2+1)+3=0\\Expand\\P^2+3-7P^2-7+3=0\\-6P^2-1=0\\6P^2=-1\\p^2=-1/6\\P=\sqrt{-1/6}[/tex]
Since:
[tex]P=2^x\\2^x=\sqrt{-1/6}\\xlog2=log(\sqrt{-1/6}) \\x = \frac{log\sqrt{-1/6}}{log2}[/tex]
