Question

Solution 2^2x+3-7(2^2x+1)+3=0 introduce Log​

Answer 1

Answer:

$x = \frac{log\sqrt{-1/6}}{log2}$

Step-by-step explanation:

Given the expression

$2^{2x}+3-7(2^{2x}+1)+3=0$

Let $P=2^x$

Substituting into the expression, we will have:

$P^2+3-7(P^2+1)+3=0\\Expand\\P^2+3-7P^2-7+3=0\\-6P^2-1=0\\6P^2=-1\\p^2=-1/6\\P=\sqrt{-1/6}$

Since:

$P=2^x\\2^x=\sqrt{-1/6}\\xlog2=log(\sqrt{-1/6}) \\x = \frac{log\sqrt{-1/6}}{log2}$