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Algebra Baldor
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Solutions étrangères.

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Problems Solved

$\frac{1}{x^2-x}=\frac{1}{x-1}$
$\frac{x+1}{x-1}-3=\frac{2-x}{x}$
$\log_2\left(2x\right)-\log_2\left(x+1\right)=2$
$\log_3\left(x\right)-\frac{1}{2}\log_3\left(9x\right)=\frac{1}{2}$
$\log_{13}\left(n-9\right)=\log_{13}\left(-2n-6\right)$
$\log_2\left(16\right)+3\log_3\left(y\right)=12-\log_3\left(y\right)$
$\log_2\left(3x-1\right)-\log_2\left(x+2\right)=4$

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