Solve the inequality for $x$: $\log_4 (x^2 − 2x + 1) < \log_2 3$ Solve the inequality for $x$: $$\log_4 (x^2 − 2x + 1) < \log_2 3$$
I got two answers and I'm not sure if I did it correctly. 
1st ans: $(-2,1)\cup (1,4)$ 
2nd ans: $x \neq -2,4$
 A: Indeed, $$\begin{align}\log_4(x^2-2x+1)&<\log_2(3)\\\log_4((x-1)^2)&<\log_2(3)\\2\log_4(|x-1|)&<\log_2(3)\\\log_2(|x-1|)&<\log_2(3)\qquad\text{since $2=\log_2(4)$}\end{align}$$
hence, $|x-1|<3$ since $\log_2$ is increasing and one to one. Thus, $-3<x-1<3\implies -2<x<4$. But of course, $x\neq 1$ since $\log_2(0)$ is undefined. Hence, the interval for $x$ is $(-2,1)\cup(1,4)$.
A: \begin{array}{c}
   \log_4 (x^2 − 2x + 1) < \log_2 3 \\
   x^2 − 2x + 1 < 4^{\log_2 3} \\
   0<(x-1)^2 < 2^{2\log_2 3} \\
   0<(x-1)^2 < 2^{\log_2 3^2} \\
   0<(x-1)^2 < 3^2 \\
   \text{$-3 < x-1 < 3$ and $x \ne 1$}  \\
   \text{$-2 < x < 4$ and $x \ne 1$} \\
   x \in (-2,1) \cup (1,4)
\end{array}
A: By the change of base formula, $\log_4{x^2-2x+1}$ is equal to $\frac{\log_2{x^2-2x+1}}{\log_2{4}}$, or $\frac{1}{2} \log_2{x^2-2x+1}$. Since both sides are in the same base, we can directly solve the inequality:
$$\frac{1}{2} \log_2{x^2-2x+1} < \log_2{3}$$
$$\log_2{x^2-2x+1} < 2 \log_2{3}$$
$$x^2-2x+1 < (2^{\log_2{3}})^2$$
$$x^2-2x+1<9$$
$$x^2-2x-8<0$$
Now factorise the quadratic equation to get its roots, then set up the inequality. For the second question, when does $\log_4{x^2+2x+1}$ equal $0$? How can you apply this constraint to your inequality?
A: With the use of $$\log_4 (x^2 − 2x + 1) =\frac{\log_2 (x^2 − 2x + 1)}{\log_2 4},$$ rewrite the given inequality as
$$ \log_2 (x^2 − 2x + 1) < 2\log_2 3$$ 
or 
$$ \log_2 (x^2 − 2x + 1) < \log_2 9.$$ 
The function is increasing, thus this holds iff 
$$0<(x^2 − 2x + 1) <9.$$ This gives the set of solutions $x \in(-2,1)\cup(1,4).$
