$\log\left(x\right)=\log\left(x+3\right)-\log\left(x-1\right)$
$\frac{x^2}{x^3}+x$
$-5-12+8-6+4-3$
$-8\cdot\left(x-1\right)^2\cdot\left(x+3\right)\cdot\left(2x-\frac{1}{3}\right)=0$
$\left(10-3x\right)^5$
$4+3x=13$
$66-\left(0.042q+27.1\right)$
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