If x is satisfied the inequality #log_(x+3)(x^2-x) < 1#, the x may belongs to the set?

A) #x in (-3,-2)#
B) #x in (-1,3)#
C) #x in (1,3)#
D) #x in (-1,0)#
The answer is A), C) and D) for your reference

1 Answer
Noah G
Aug 19, 2017

I got #x in (-3, -2) uu (-1, 0) uu (0, 3)#, so #A, B and D#

Explanation:

By the definition of the logarithm, we have:

#x^2 - x < (x + 3)^1#

#x^2 - x < x + 3#

#x^2 - 2x - 3 < 0#

Solving like an equation:

#x^2 - 2x - 3 = 0#

#(x- 3)(x + 1) = 0#

#x= 3 or -1#

If we select a test point, say #x = 0#, we get:

#0^2 - 2(0) - 3 < 0 color(green)(√)#

However, if we use #x = -2#, we realize it doesn't work. So our solution set is #x in (-1, 3)#. This is only for the quadratic though. We haven't considered the the number in #x^2 - x# must be positive for the logarithm to be defined. Hence, we have:

#x^2 - x > 0#

#x^2 - x = 0#

#x(x - 1) = 0#

#x = 0 or 1#

If we repeat the process with test points, we realize that the solution is #(-oo, 0)# and #(1, oo)#. Therefore, we must eliminate #(0, 1)# from our solution set above.

Now, we must also guarantee that #x + 3 > 0#. Then #x > -3#. This means that we can include #x > -3# in our solution set. The set becomes #(-3, 3)#. However, if you try #x = -1.5#, (or any number in the interval #(-2, -1)#, you will realize that the result you get is not within the range of the problem, that' s to say it will be greater, instead of less than #1#.

The answer is therefore :

#x in (-3, -2) uu (-1, 0) uu (0, 3)#

A graphical verification yields the same results.

Hopefully this helps!

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