Sol. The given inequality is

log a > 1

or > 1

Let > 1, i.e. x 2 < 9. In this case the given inequality is equivalent to >

this inequality may be written as x 2 + 16x – 17 < 0, i.e., –17 < x < 1 but the condition of this can (i.e. –3 < x < 3) is only satisfied by those x located in the interval – 3 < x < 1.

Let 0 < < 1 , i.e. 9 < x 2 < 25 or 3 < x < 5 and – 5 < x < –3. Here the original inequality is equivalent to the double inequality 0 < < i.e. x 2 + 2x – 24 < 0, x 2 + 16x – 17 > 0

The first inequality and 2 nd inequality of this system has the solutions are – 6 < x < 4 and (x > 1 and x < – 17) but the condition of this case (3 < x < 5 and – 5 < x –3) is only satisfied by those x located in the interval 1 < x < 4.Thus combining two cases, we have the solution of the original inequality, which consists of two intervals : –3 < x < 1 and 3 < x < 4.