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/
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