Values of $a$ such that $\log_2(x^2)=a\sqrt{\log_2(x^4)}+a-1$ has 4 solutions

Question

I'm studying logarithms and exponential functions and am working on this question:

Find all real values for $a$ such that the equation $\log_2(x^2)=a\sqrt{\log_2(x^4)}+a-1$ has exactly 4 real solutions.

Can anyone help me?

Answer 1

Put $\sqrt{\log_2(x^2)}=X$

the equation becomes

$X^2-aX\sqrt{2}-a+1=0$

it must have two real roots .

thus

$\delta=2(a^2+2a-2)>0$

and

$$a\in (-\infty,-1-\sqrt{3})\cup (-1+\sqrt{3},+\infty)$$

Answer 2

Hint

Write $t=\log_2x^2$ and note that $\log_2 x^4=\log_2 (x^2)^2=2\log_2 x^2.$ So, the original equation can be written as

$$t=a\sqrt{2t}+a-1.$$ Find $a$ such that this equation has two different solutions $t_1,t_2.$ Then, $$\log_2x^2=t_i\iff x^2=2^{t_i} \iff x=\pm\sqrt{2^{t_i}}$$ gives you the four different solutions.

Answer 3

You can write:

\begin{equation} 2\log_2x-2a\sqrt{\log_2x}+a-1=0\\2t^2-2at+a-1=0 \end{equation}

where $t=\log_2x$ and you have to solve this equation enforcing $\Delta>0$.