Answer :

The equation is given to be:

[tex]\log _{\frac{1}{3}}6xy[/tex]

Recall the logarithm rule:

[tex]\log _{\frac{1}{a}}\mleft(x\mright)=-\log _a\mleft(x\mright)[/tex]

Therefore, the expression becomes:

[tex]\log _{\frac{1}{3}}6xy=-\log _36xy[/tex]

Factorize the number 6:

[tex]\begin{gathered} 6=2\cdot3 \\ \therefore \\ -\log _36xy=-\log _3(3\cdot2xy) \end{gathered}[/tex]

Recall the rule of logarithm:

[tex]\log _c\mleft(ab\mright)=\log _c\mleft(a\mright)+\log _c\mleft(b\mright)[/tex]

Thus, the expression becomes:

[tex]-\log _3(3\cdot2xy)=-(\log _33+\log _32+\log _3x+\log _3y)[/tex]

Recall the rule:

[tex]\log _aa=1[/tex]

Hence, the expression simplifies to give:

[tex]-(\log _33+\log _32+\log _3x+\log _3y)=-(1+\log _32+\log _3x+\log _3y)[/tex]

Expanding, we have the answer to be:

[tex]\Rightarrow-1-\log _32-\log _3x-\log _3y[/tex]

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