Solving a difficult logarithmic inequality involving a fraction

Question

My question regards solving the following logarithmic inequality for x: $$\dfrac{\log_{2} (x^{2}-6x+8)}{\log_{2} (x-8)}< 1.$$ I have become very confused as to how to solve more complicated logarithmic inequalities like these.

As far as my attempts go, I have recognized that the following domain restrictions apply: $$x^{2}-6x+8 > 0,$$ $$x-8 > 0,$$ $$\log_{2} (x-8) \ne 0.$$ Thus, it follows from further calculation and intersection of these domains that the domain of the original inequality is $$(8, 9) \cup (9, \infty).$$ However, I see that the final answer is simply $(8, 9)$. What step have I missed? Have I made an arithmetic mistake? Many thanks for any advice or direction.

Answer 1

Your expression is not defined when $x<8.$ Can't take the log of a negative number.

And it is not defined when $x = 9,$ Can't divide by $0.$

Break it into cases.

$8<x<9$ numerator is positive, denominator is negative, the inequality is true.

$x>9, x^2 - 6x + 8 > x + 8$ for all $x$ is in this case, and $\log (x^2 - 6x + 8) > \log (x + 8)$ and the inequality is not true.

your solution set is $(8,9)$

Answer 2

Hint: Use the change of base rule in reverse

$$\frac{\log_2(x^2-6x+8)}{\log_2(x-8)}=\log_{(x-8)}(x^2-6x+8)$$

Then undo the $\log$s.

Answer 3

The domain of this inequation is correct. You have two cases

• $8<x<9$. Then $\;\log_2(x-8)<\log_2(9-8)=0$, and $x^2-6x+8=(x-3)^2-1>24>0$. Hence the l.h.s. is negative, and it's certainly $<1$.
• $x>9$: both logs are positive, so the inequation is equivalent to $\;\log_2(x^2-6x+8)<\log_2(x-8)$, i.e. to $\;x^2-6x+8<x-8\iff x^2-7x+16<0$. The discriminant is negative, so it has no solution.