Question

Question #cd974

Answer 1

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

Explanation:

We have:

`log(x - 1)(8 - x) < 1`

`log(-x^2 - 8 + x + 8x) < 1`

`log(-x^2 + 9x - 8) < 1`

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

`-x^2 + 9x - 8 < 10^1`

`0 < x^2 - 9x + 18`

Solve as an equation and use test points.

`0 = x^2 - 9x + 18`

`0 = (x -6)(x - 3)`

`x= 6 or 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!