# Question #cd974

Feb 4, 2018

The solution is $1 < x < 3$ and $6 < x < 8$

#### Explanation:

We have:

$\log \left(x - 1\right) \left(8 - x\right) < 1$

$\log \left(- {x}^{2} - 8 + x + 8 x\right) < 1$

$\log \left(- {x}^{2} + 9 x - 8\right) < 1$

Since the log is assumed to be base $10$:

$- {x}^{2} + 9 x - 8 < {10}^{1}$

$0 < {x}^{2} - 9 x + 18$

Solve as an equation and use test points.

$0 = {x}^{2} - 9 x + 18$

$0 = \left(x - 6\right) \left(x - 3\right)$

$x = 6 \mathmr{and} x = 3$

Clearly $x = 0$ is a solution therefore $x < 3$ and $x > 6$ is a solution. However, due to the restrictions on the original log, we cannot have $x > 8$ or $x < 1$. Thus, $1 < x < 8$, but this isn't true in the interval $3 < x < 6$, thus our solution intervals are $1 < x < 3$ and $6 < x < 8$

Hopefully this helps!