1/(1+log(x)) + 1/(1-log(x)) > 2 solve for x

Question

i have been stuck on  this problem lately 

Answer 1

Hello Yash,

This problem is rather tricky. I'm not sure what level of math you're at - if it's precalculus or above then it may make sense for you to do this using algebra. If you're lower than that they may expect you to do it using a graphing calculator. Below I'll explain how to do it using algebra.

 

Also, I'll assume "log" means log₁₀.

 

Generally when you solve inequalities beyond basic (linear) inequalities, the procedure is:

1. Get everything on one side (say, the left) so you have a zero on the right, and simplify the left side.

2. Set the numerator and the denominator each equal to zero, and solve for x. (If the numerator is zero, the whole expression equals zero. If the denominator is zero, it's undefined.)

3. Write a number line and mark all the places you found in part (2). Then take a number in each interval and plug it into the expression on the left to see whether it's positive or negative.

 

Here it goes:

First, get everything on the left side and simplify. We get:

 

1/(1+logx) + 1/(1-logx) - 2 > 0

 

((1 - logx) + (1 + logx) - 2(1 + logx)(1 - logx)) / ((1 + logx)(1 - logx)) > 0   (here I got a common denominator)

 

((2 - 2(1-log²x)) / (1 - log²x)) > 0   (simplified)

 

2log²x / (1 - log²x) > 0  (simplified more)

 

 

Second, set top and bottom equal to zero.

 

Numerator:  2log²x = 0  ⇒  This only happens if logx = 0, that is, x = 1.

 

Denominator:  1 - log²x = 0  ⇒  This happens if logx = 1 or -1, so x = 10 or 1/10.   (again assuming log means log₁₀)

 

 

Third, write your number line. Usually you write a number line with arrows going to the left and right, but in this case, since log is undefined for x ≤ 0, the arrow just starts at 0 and goes to the right.

 

Mark your three points: 1/10, 1, and 10. Then test each interval. The numerator is positive on each interval, so you can actually just test the denominator. You will find that it is positive in the middle two intervals, and negative in the outer two.

 

Therefore, your answer is (1/10, 1) U (1, 10).