Question

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

Answer 1

`x = (log√(-1/6))/(log2)`

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=√(-1/6)`

Since:

`P=2^x\\2^x=√(-1/6)\\xlog2=log(√(-1/6)) \\x = (log√(-1/6))/(log2)`