Solving an Inequality

Date: 6/17/96 at 10:27:21
From: Novica
Subject: Solving an inequality

Solve inequation:

    6/(2X+1) > (1/X)(1+log2(2+X)).

log2 stands for logarithm of base 2.

Date: 6/28/96 at 14:22:26
From: Doctor Jerry
Subject: Re: Solving an inequality

I assume you know that log2(w) = ln(w)/ln(2), where ln is natural 
logarithm. If this is something you want to discuss, please let me 
know.

Otherwise, we must assume that X > -2, so that log2 or ln are defined.  
Also, we must avoid X = -1/2 and X = 0.  These restrictions divide the 
axis into three regions: -2 < X < -1/2, -1/2 < X < 0, and X > 0. 

By graphing the function 6/(2X+1) - (1/X)(1+ln(2+X)/ln(2)) in each of 
these intervals, it is easy to see that the inequality is true only in 
the interval -1/2 < X < 0.

I tried for a while to give a non-graphical solution, using 
substitution, but I was unable to find a simpler equation.  I thought 
about using power series, but decided it would be quite messy.

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/