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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: 7,649 Thanks: 2630 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 half-past one in the morning, I'm going to bed instead. I'd be very surprised if that algebra is without errors. Thanks from Olinguito Last edited by v8archie; April 12th, 2014 at 10:39 PM.
April 13th, 2014, 05:55 AM   #3
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Quote:
 Originally Posted by v8archie This we can solve for a^2, but it's half-past one in the morning, I'm going to bed instead. I'd be very surprised if that algebra is without errors.
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: 7,649 Thanks: 2630 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
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Posts: 425
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Quote:
 Originally Posted by v8archie 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."
Any equation of any degree $\displaystyle a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0=0$ can have real solutions and/or complex and so and any inequality of this type,because any inequality can turn into an equation.

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