Krdytkyu

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: $x neq -2,4$

Indeed, $$begin{align}log_4(x^2-2x+1)&<log_2(3)\log_4((x-1)^2)&<log_2(3)\2log_4(|x-1|)&<log_2(3)\log_2(|x-1|)&<log_2(3)qquadtext{since $2=log_2(4)$}end{align}$$
hence, $|x-1|<3$ since $log_2$ is increasing and one to one. Thus, $-3<x-1<3implies -2<x<4$. But of course, $xneq 1$ since $log_2(0)$ is undefined. Hence, the interval for $x$ is $(-2,1)cup(1,4)$.

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^{2log_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}

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?

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
$$ log_2 (x^2 − 2x + 1) < 2log_2 3$$
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 $x in(-2,1)cup(1,4).$