Logarithmic inequality

Question

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?

Answer 1

A start: We have $\log_8(x^2-4x+3)\lt 1$ precisely if $0\lt x^2-4x+3\lt 8^1$.

Answer 2

Here's are some steps $$ \log_8 (x^2-4x+3)\lt 1 $$ $$ 8^{\log_8 (x^2-4x+3)}\lt 8^1 $$ $$ 0\lt x^2-4x+3\lt 8 $$ $$ -8\lt x^2-4x-5\lt 0 $$ $$ -8\lt (x+1)(x-5)\lt 0 $$ Can you take it from here?

Answer 3

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, $(1)\cap(2) \implies x\in(-1,1)\cup(3,5)$