The value of the determinant- -log x log y log z log 2x log 2y log 2z log 3x log 3y log 3z is

Δ = log x amp;log y amp;log z log 2x amp;log 2y amp;log 2z log 3x amp;log 3y amp;log 3z Δ = log x amp;log y amp;log z log 2+l ...

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Evaluation of a Determinant

The value of ...

Question

The value of the determinant
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} l o g x & l o g y & l o g z l o g 2 x & l o g 2 y & l o g 2 z l o g 3 x & l o g 3 y & l o g 3 z \end{matrix} ∣ ∣ ∣ ∣$ is

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log (xyz)

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log (6 xyz)

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6 log (xyz)

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Solution

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Similar questions

⎛ ⎜ ⎝ \begin{matrix} log x & log y & log z log 2 x & log 2 y & log 2 z log 3 x & log 3 y & log 3 z \end{matrix} ∣ ∣ ∣ ∣ ∣

⎛ ⎜ ⎝ \begin{matrix} log x & log y & log z log 2 x & log 2 y & log 2 z log 3 x & log 3 y & log 3 z \end{matrix} ∣ ∣ ∣ ∣ ∣

x, y

z

∣ ∣ ∣ ∣ \begin{matrix} 1 & log, y & log, z log, x & 1 & log, z log, x & log, y & 1 \end{matrix} ∣ ∣ ∣ ∣

Δ = ∣ ∣ ∣ ∣ \begin{matrix} log x log 2 x log 3 x \end{matrix} \begin{matrix} log y log 2 y log 3 y \end{matrix} \begin{matrix} log z log 2 z log 3 z \end{matrix} ∣ ∣ ∣ ∣

x \frac{d y}{d x} = y (l o g y - l o g x + 1)

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