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Question

Mathematics Question on Determinants

Evaluate the determinant Δ=log3512log43 log38log49\Delta=\left|\begin{matrix}log_{3} \,512&log_{4} \,3\\\ log_{3} \,8 &log_{4} \,9\end{matrix}\right|

A

15/215/2

B

1212

C

14/314/3

D

66

Answer

15/215/2

Explanation

Solution

We have, Δ=log3512log43 log38log49Δ=log329log223 log323log2232 \Delta=\left|\begin{matrix}log_{3} \,512&log_{4} \,3\\\ log_{3}\, 8&log_{4}\, 9\end{matrix}\right| \Rightarrow \Delta=\left|\begin{matrix}log_{3} \, 2^{9}&log_{2^{2}} \,3\\\ log_{3} \,2^{3}&log_{2^{2}} \, 3^{2} \end{matrix}\right| Δ=9log3212log23 3log3222log23\Rightarrow \Delta=\left|\begin{matrix}9\,log_{3} \,2&\frac{1}{2}log_{2} \,3\\\ 3 log_{3} \,2&\frac{2}{2} log_{2} \,3\end{matrix}\right| [logapmn=nplogam]\left[\because\quad log_{a^{p}}\, m^{n}=\frac{n}{p} log_{a}\, m\right] Δ=(9log32)×(log23)(12log23)(3log32)\Rightarrow \Delta=\left(9\, log_{3} \,2\right)\times\left(log_{2} \,3\right) -\left(\frac{1}{2}\,log_{2}\, 3\right)\left(3 \,log_{3} \,2\right) Δ=9(log32×log23)32(log23×log32)\Rightarrow \Delta=9 \left(log_{3} \,2\times log_{2}\, 3\right)-\frac{3}{2}\left(log_{2}\, 3\times log_{3} \,2\right) Δ=932Δ=152\Rightarrow \Delta=9-\frac{3}{2} \Rightarrow \Delta =\frac{15}{2} [logba×logab=1]\quad\left[\because log_{b} \,a\times log_{a}\, b=1\right]