15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T485,

The given determinant is

$$ \left|\begin{array}{ccc} 1 & \log _a b & \log _a c \\ \log _b a & 1 & \log _b c \\ \log _c a & \log _c b & 1 \end{array}\right| $$

Using the change of base formula $\log _x y=\frac{\ln y}{\ln x}$ then determinant becomes.

$$ \left|\begin{array}{ccc} 1 & \frac{\ln b}{\ln a} & \frac{\ln c}{\ln a} \\ \frac{\ln a}{\ln b} & 1 & \frac{\ln c}{\ln b} \\ \frac{\ln a}{\ln c} & \frac{\ln b}{\ln c} & 1 \end{array}\right| $$

Let $X=\ln a, Y=\ln b, Z=\ln c$. the determinant is

$$ \left|\begin{array}{ccc} 1 & Y / X & Z / X \\ X / Y & 1 & Z / Y \\ X / Z & Y / Z & 1 \end{array}\right| $$

$$ \begin{aligned} &\text { Expand the determinant }\\ &\begin{aligned} \left.1-\frac{Z}{Y} \cdot \frac{Y}{Z}\right)-\frac{Y}{X}\left(\frac{X}{Y}-\frac{Z}{Y}\right. & \left.\cdot \frac{X}{Z}\right) +\frac{Z}{X}\left(\frac{X}{Z} \cdot \frac{Y}{Y}-1 \cdot \frac{X}{Z}\right) \end{aligned} \end{aligned} $$

$$ \begin{aligned} & =1(1-1)-\frac{Y}{X}\left(\frac{X}{Y}-\frac{X}{Y}\right)+\frac{Z}{X}\left(\frac{X}{Z}-\frac{X}{Z}\right) \\ & =1(0)-\frac{Y}{X}(0)+\frac{Z}{X}(0) \\ & =0 \end{aligned} $$

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