<s>[INST] Logarithmic inequality Solve the inequality:
$$
\log_8(x^2-4x+3) < 1
$$
$$
\log_8(x^2-4x+3) < \log_8(8)
$$
$$
\log_8(x^2-4x+3) - \log_8(8) <0
$$
$$
\log_8 [(x^2-4x+3)/8] < 0
$$
Thats what I did for the question so far... and I'm confused as to what to do next. Can someone verify this for me?
 [/INST] Firstly, $\log(n)$ is defined only if $n \in (0,\infty)$
$$
\begin{align}
\Rightarrow\quad x^2 - 4x + 3 &\in (0,\infty)\\
\Rightarrow\quad x^2 - 4x + 3 &> 0\\
\Rightarrow (x-3)(x-1) &> 0\\
\end{align}
$$
$$
\boxed{\therefore\quad x \in (-\infty, 1)\cup(3,\infty) 
\quad\text{or}\quad x\in \mathbb R \sim [1,3]}\tag{1}
$$
Now, $\log_b(n) = y \iff n = b^y$,
$$
\begin{align}
\log_8(x^2-4x+3) &< 1\\
\Rightarrow x^2 - 4x + 3 &< 8^1\\
\Rightarrow x^2 - 4x - 5 &< 0\\
\Rightarrow (x+1)(x-5) &< 0
\end{align}
$$
$$
\boxed{\therefore x\in\left(-1,5\right)} \tag{2}
$$
Remember that both $(1)$ and $(2)$ have to be satisfied at the same time.
Now, $(1)\cap(2) \implies x\in(-1,1)\cup(3,5)$
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