# Solve exponential-polynomial equation Solve the equation in <mrow class="MJX-TeXAtom-ORD">

Solve exponential-polynomial equation
Solve the equation in $\mathbb{R}$
${10}^{-3}{x}^{{\mathrm{log}}_{10}x}+x\left({\mathrm{log}}_{10}^{2}x-2{\mathrm{log}}_{10}x\right)={x}^{2}+3x$
To be fair I wasn't able to make any progress. I tried using substitution for the logarithms, but it didn't help at all.
This is a contest problem, so there should be a nice solution.
Any help? Clue?
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Isla Klein
Clearly $x>0$. Let $x={10}^{y}$. Then we have ${10}^{-3}{x}^{y}+x\left({y}^{2}-2y\right)={x}^{2}+3x$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{10}^{-3}{x}^{y-1}+\left(y-3\right)\left(y+1\right)=x$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{10}^{{y}^{2}-y-3}+\left(y-3\right)\left(y+1\right)={10}^{y}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\underset{⏟}{{10}^{y}\left({10}^{\left(y-3\right)\left(y+1\right)}-1\right)}+\underset{⏟}{\left(y-3\right)\left(y+1\right)}=0$
Now note that if $\left(y-3\right)\left(y+1\right)>0$, both terms on the left are positive, hence the equation cannot have a solution. Similarly, if $\left(y-3\right)\left(y+1\right)<0$, both terms are negative and again we cannot have a solution. Hence the only solutions are when $\left(y-3\right)\left(y+1\right)=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y\in \left\{-1,3\right\}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\in \left\{\frac{1}{10},1000\right\}$
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Raul Walker
that is not a hard question !
you can assume :
${10}^{a}=x$
then your problem turn out to be :
${x}^{2}+\left(3-{a}^{2}+2a\right)x-\frac{{x}^{a}}{{10}^{3}}=0$