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April 12th, 2014, 09:19 PM  #1 
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  Another inequality
To solve the inequality $\displaystyle x^4+log_23^{(x^2+4)}+5\leq0$.

April 12th, 2014, 10:27 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
\begin{align*} x^4+log_23^{(x^2+4)}+5 &\leq 0 \\ x^4+\left( x^2 + 4 \right) \frac{\log{3}}{\log{2}} + 5 &\leq 0 \\ x^4 + \frac{\log{3}}{\log{2}} x^2 + \left(\frac{4\log{3}}{\log{2}} + 5\right) &\leq 0 \\ x^2 = \frac{1}{2}\left( \frac{\log{3}}{\log{2}} \pm \sqrt{ \frac{\log^2{3}}{\log^2{2}}  4\left(\frac{4\log{3}}{\log{2}} + 5\right) } \right) \end{align*} I reckon we have no real roots here, so I suppose that this is another one where we want to make the imaginary part of the equation disappear for complex $x$. Thus $x = a +b\imath$ \begin{align*} \operatorname{Im}{\left(x^4 + \frac{\log{3}}{\log{2}} x^2\right)} &= 0 \\ \operatorname{Im}{\left(a^4 + 4a^3b\imath  6a^2b^2 4ab^3\imath + b^4 + \frac{\log{3}}{\log{2}} \left(a^2 + 2ab\imath  b^2 \right) \right)} &= 0 \\ 4a^3b 4ab^3 + \frac{\log{3}}{\log{2}} 2ab &= 0 \\ \end{align*} Which gives $b = 0$ (which we have already dismissed) or $$ b^2 = a^2 + \frac{\log{3}}{2\log{2}} $$ Which we substitute into \begin{align*} \operatorname{Re}{\left(a^4 + 4a^3b\imath  6a^2b^2 4ab^3\imath + b^4 + \frac{\log{3}}{\log{2}} \left(a^2 + 2ab\imath  b^2 \right) \right)} &\leq \left(\frac{4\log{3}}{\log{2}} + 5\right) \\ a^4  6a^2b^2 + b^4 + \frac{\log{3}}{\log{2}} \left(a^2  b^2 \right) &\leq \left(\frac{4\log{3}}{\log{2}} + 5\right) \\ a^4  6a^2\left(a^2 + \frac{\log{3}}{2\log{2}}\right) + \left(a^2 + \frac{\log{3}}{2\log{2}}\right)^2 + \frac{\log{3}}{\log{2}} \left(a^2  \left(a^2 + \frac{\log{3}}{2\log{2}}\right) \right) &\leq \left(\frac{4\log{3}}{\log{2}} + 5\right) \\ 4a^4 \frac{2\log{3}}{\log{2}}a^2  \frac{\log^2{3}}{4\log^2{2}} &\leq \left(\frac{4\log{3}}{\log{2}} + 5\right) \\ 4a^4 + \frac{2\log{3}}{\log{2}}a^2 + \frac{\log^2{3}}{4\log^2{2}} \frac{4\log{3}}{\log{2}}  5 &\ge 0 \\ \end{align*} This we can solve for a^2, but it's halfpast one in the morning, I'm going to bed instead. I'd be very surprised if that algebra is without errors. Last edited by v8archie; April 12th, 2014 at 10:39 PM. 
April 13th, 2014, 05:55 AM  #3 
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  I do not understand!For what values of $\displaystyle x$ is satisfied the inequality?
Last edited by Dacu; April 13th, 2014 at 06:28 AM. 
April 13th, 2014, 06:40 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
I didn't go as far as getting the values of $x$ because the algebra is getting far too horrible. Suffice to say that $x$ will be complex, satisfying $x = a + b\imath$ where $a$ and $b$ satisfy equations given in my post above. If my efforts are correct, the question is phrased badly and should say something like "Find values of $z \in \mathbb{C}$ such that $z^4+log_23^{(z^2+4)}+5$ is real and less than or equal to zero." 
April 13th, 2014, 08:47 PM  #5  
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  Quote:
 

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