I have a problem in which I need to solve for $x$ in the question (in image) below, but I don't even understand it. Is this suppose to be a matrix? $$\begin{vmatrix}\ln(2x+1)&2-\ln x\\1&1\end{vmatrix}=0$$
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$\begingroup$ I think a picture or rendering it in LaTeX would prove more useful, because the bit you pasted renders as nonsense for me. $\endgroup$ – Eevee Trainer Nov 29 '20 at 9:26
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$\begingroup$ @EeveeTrainer I've added a picture! $\endgroup$ – Mya Ishikawa Nov 29 '20 at 9:30
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4$\begingroup$ This looks like the determinant of a $2 \times 2$ matrix by my eye. $\endgroup$ – Eevee Trainer Nov 29 '20 at 9:32
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The bar notation means taking the determinant of the matrix enclosed. This is a shorthand for $\det M$.
answered Nov 29 '20 at 9:32
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The bar notation is used to indicate the determinant of a matrix. It converts a matrix into a scalar. For example, $$\begin{align}\begin{vmatrix} a & b\\c & d \end{vmatrix}=ad - bc \end{align}$$ Other common notations are $\det A$ and $\det (A)$.
In particular, for your case, the mentioned determinant becomes $$\begin{align}\begin{vmatrix} \ln (2x+1) & 2-\ln x\\1 & 1 \end{vmatrix}=\ln (2x+1)-2+\ln x = \ln (2x^2+x)-2 \end{align}$$
Hope this helps :)
