<s>[INST] 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: $x \neq -2,4$
 [/INST] 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?
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