If log 1+x square to the base 2x square = 1 then x=?
Answers
Answer:
Given x satisfies the inequality log
x+3
(x
2
−x)<1
Consider log
x+3
(x
2
−x)<1
By definition of logarithm, we have
x
2
−x<(x+3)
1
⇒x
2
−x<x+3
⇒x
2
−2x−3<0
Solving it as quadratic equation, x
2
−2x−3=0, we get (x−3)(x+1)=0
⇒x=3or−1
So our solution set is x∈(−1,3)
This is only for quadratic though. We haven't considered the number in x
2
−x must be positive for the logarithm to be denied.
Hence we have, x
2
−x>0
⇒x
2
−x=0
⇒x(x−1)=0
⇒x=0or1
If we repeat the process with test points, we realize that the solution is (−∞,0) and (1,∞). Therefore, we must eliminate (0,1) from our solution set above.
Now, we must also guarantee that x+3>0.
⇒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 (for 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).
Therefore x∈(−3,−2)∪(−1,0)∪(1,3)
A graphical verification yields the same results