find the range of values of x

**Punch** · August 24th 2012, 07:39 PM

Find in terms of a, the range of x that satisfy the inequality $ln(2x-\frac{a}{x})=>0$

$2x-\frac{a}{x}=>1$

$\frac{2x^2-x-a}{x}=>0$

**Prove It** · August 24th 2012, 09:02 PM

Originally Posted by Punch

Find in terms of a, the range of x that satisfy the inequality $ln(2x-\frac{a}{x})=>0$

$2x-\frac{a}{x}=>1$

$\frac{2x^2-x-a}{x}=>0$

First of all, you need to note that a logarithm is only defined for positive values, so $\displaystyle \begin{align*} 2x - \frac{a}{x} > 0 \end{align*}$ . This requires that both $\displaystyle \begin{align*} 2x \end{align*}$ and $\displaystyle \begin{align*} \frac{a}{x} \end{align*}$ are positive, which means $\displaystyle \begin{align*} x \end{align*}$ and $\displaystyle \begin{align*} a \end{align*}$ are also both positive. Then we have

$\displaystyle \begin{align*} 2x &> \frac{a}{x} \\ 2x^2 &> a \textrm{ since we know } 2x > 0 \implies x > 0 \\ x^2 &> \frac{a}{2} \\ x &> \sqrt{\frac{a}{2}} \\ x &> \frac{\sqrt{a}}{\sqrt{2}} \\ x &> \frac{\sqrt{2a}}{2} \end{align*}$

Now solving the original inequality

$\displaystyle \begin{align*} \ln{\left(2x - \frac{a}{x}\right)} &\geq 0 \\ 2x - \frac{a}{x} &\geq 1 \\ \frac{2x^2 - x - a}{x} &\geq 0 \\ 2x^2 - x - a &\geq 0 \textrm{ since we already know } x > 0 \\ 2x^2 - x &\geq a \\ x^2 - \frac{1}{2}x &\geq \frac{a}{2} \\ x^2 - \frac{1}{2}x + \left(-\frac{1}{4}\right)^2 &\geq \frac{a}{2} + \left(-\frac{1}{4}\right)^2 \\ \left(x - \frac{1}{4}\right)^2 &\geq \frac{8a}{16} + \frac{1}{16} \\ \left(x - \frac{1}{4}\right)^2 &\geq \frac{8a + 1}{16} \\ \left| x - \frac{1}{4} \right| &\geq \frac{\sqrt{8a + 1}}{4} \\ x - \frac{1}{4} \leq -\frac{\sqrt{8a + 1}}{4} \textrm{ or } x - \frac{1}{4} &\geq \frac{\sqrt{8a + 1}}{4} \\ x \leq \frac{1 - \sqrt{8a + 1}}{4} \textrm{ or } x &\geq \frac{1 + \sqrt{8a + 1}}{4} \end{align*}$

So finally, the solution of this inequality is the intersection of these two sets...

$\displaystyle \begin{align*} x \in \left( \frac{\sqrt{2a}}{2} , \infty \right) \cap \left\{ \left(- \infty , \frac{1 - \sqrt{8a + 1}}{4} \right] \cup \left[ \frac{1 + \sqrt{8a + 1}}{4}, \infty \right) \right\} \end{align*}$

**Punch** · August 25th 2012, 05:58 AM

I agree that $2x-\frac{a}{x}>0$ but i do not agree that this requires both $2x$ and $\frac{a}{x}$ to be positive.

If I sub x=-0.5 and taking a=1, $2(-0.5)-\frac{1}{-0.5}=1>0$

**Prove It** · August 25th 2012, 06:06 AM

Originally Posted by Punch

I agree that $2x-\frac{a}{x}>0$ but i do not agree that this requires both $2x$ and $\frac{a}{x}$ to be positive.

If I sub x=-0.5 and taking a=1, $2(-0.5)-\frac{1}{-0.5}=1>0$

I stand corrected.

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