Leeland C. answered 05/18/14
Tutor
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Mathematics and English expert through High School level classes
In this equation, you have a natural log of some number. This is the same as log base e of that number. Since e is not 0, the only limitation on x is that (x^2-1x-20) cannot be less than or equal to zero, since e to any power will never equal 0 or be negative.
All we need to do here is set what is in the parentheses equal to 0 to find what x cannot equal, then determine where x would also be negative.
0=x^2-x-20 (1x is the same thing as x)
0=(x-5)(x+4) (I unFOILed this by considering possible numbers that multiply to equal 20 [1,20; 2,10; 4,5] and then found the combination that would equal -1 when one was negative and one was positive)
From this, we see that if x=5 or x=-4, the equation will be true. (5-5)(5+4)=0*9=0 (5-(-4))(-4+4)=9*0=0
So we know that x CANNOT equal either 5 or -4. Now we need to test to see where it would be negative. Use the steps from the previous problem to test this. For simplicity, I'm just going to state that x would be negative between -4 and 5. So interval notation of the answer would look like this:
(-∞,-4) U (5,∞)
Dalia S.
05/18/14