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How to solve the equation $$\mathrm{e}^x + (x^3-x)\cdot \ln(x^2+x+2) - \mathrm{e}^{\sqrt[3]{x}}=0?$$

I tried. We have $x=1$ is a root of the equation. If $x>1$, $x > \sqrt[3]{x}$, therefore $\mathrm{e}^x>\mathrm{e}^{\sqrt[3]{x}}$ and $x^3 >x$, $\ln(x^2+x+2) > \ln4>0$. Then $$\mathrm{e}^x + (x^3-x)\cdot \ln(x^2+x+2) - \mathrm{e}^{\sqrt[3]{x}}> 0.$$ How about $x < 1$?

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First, let us reorder the addends in this way: $$ e^x - e^{\sqrt[3]x} + (x^3−x)\ln(x^2+x+2) = 0 $$ Note that $x^3-x=0$ iff $x-\sqrt[3]x=0$ iff $x=0,\pm1$ and in any of these cases the equation holds.

Now, observe that $x^3-x>0$ iff $x-\sqrt[3]x>0$ iff $e^x-e^{\sqrt[3]x}>0$. Besides $x^2+x+2>1$ $\forall x$ and consequently $\ln(x^2+x+2)>0$ $\forall x$.

If $x\neq 0,\pm1$, then we can divide by $(x^3-x)$: $$ \frac{e^x - e^{\sqrt[3]x}}{x^3−x} + \ln(x^2+x+2) = 0 $$

Since $x-\sqrt[3]x>0$ iff $e^x-e^{\sqrt[3]x}>0$, then $\frac{e^x - e^{\sqrt[3]x}}{x^3−x}>0$ and therefore all the lhs is strictly greater than $0$.

Therefore the unique solutions are $x=0,\pm1$.

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x=0 and x=1 are obvious solutions of this equation. They seem to be the only ones.

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  • $\begingroup$ $x=-1$ is solution too. $\endgroup$
    – AndreasT
    Nov 14 '13 at 11:06
  • $\begingroup$ @AndreasT. You are right ! I missed that one. $\endgroup$ Nov 14 '13 at 11:10

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