Question #cd974

1 Answer
Noah G
Feb 4, 2018

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!

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