Answer:
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:log\dfrac{ {a}^{2} }{ab} + log\dfrac{ {b}^{2} }{bc} + log\dfrac{ {c}^{2} }{ca} = 0\qquad \: \\ \\& \qquad \:\sf \:log_{2}( \sqrt{6} ) + log_{2}\left(\sqrt{\dfrac{2}{3} } \right) = 1\\ \\& \qquad \:\sf \:log\dfrac{11}{15} + log \dfrac{35}{9} + log\dfrac{27}{77} = 0 \end{aligned}} \qquad \: \\ \\ [/tex]
Step-by-step explanation:
[tex]\large\underline{\sf{Solution-1}}[/tex]
Given expression is
[tex]\sf \: log\dfrac{ {a}^{2} }{ab} + log\dfrac{ {b}^{2} }{bc} + log\dfrac{ {c}^{2} }{ca} \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: log\dfrac{ {a}}{b} + log\dfrac{ {b}}{c} + log\dfrac{ {c}}{a} \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{ {a}}{b} \times \dfrac{ {b}}{c} \times \dfrac{ {c}}{a}\right) \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: logx + logy = log(xy) \: }} \\ \\ [/tex]
[tex]\sf \: = \: log(1) \\ \\ [/tex]
[tex]\sf \: = \: 0 \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: log\dfrac{ {a}^{2} }{ab} + log\dfrac{ {b}^{2} }{bc} + log\dfrac{ {c}^{2} }{ca} = 0 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex]\large\underline{\sf{Solution-2}}[/tex]
[tex]\sf \: log_{2}( \sqrt{6} ) + log_{2}\left(\sqrt{\dfrac{2}{3} }\right) \\ \\ [/tex]
[tex]\sf \: = \: log_{2}\left(\sqrt{6} \times \sqrt{\dfrac{2}{3} } \right) \\ \\ [/tex]
[tex]\sf \: = \: log_{2}\left(\sqrt{2} \times \sqrt{2}\right) \\ \\ [/tex]
[tex]\sf \: = \: log_{2}(2) \\ \\ [/tex]
[tex]\sf \: = \: 1 \\ \\ [/tex]
[tex]\sf\implies \sf \: log_{2}( \sqrt{6} ) + log_{2}\left(\sqrt{\dfrac{2}{3} } \right) = 1 \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-3}}[/tex]
[tex]\sf \: log\dfrac{11}{15} + log \dfrac{35}{9} + log\dfrac{27}{77} \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{11}{15} \times \dfrac{35}{9} \times \dfrac{27}{77}\right) \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{1}{15} \times \dfrac{35}{9} \times \dfrac{27}{7}\right) \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{1}{15} \times \dfrac{35}{1} \times \dfrac{3}{7}\right) \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{1}{5} \times \dfrac{35}{1} \times \dfrac{1}{7}\right) \\ \\ [/tex]
[tex]\sf \: = \: log\left(1\right) \\ \\ [/tex]
[tex]\sf\implies \sf \: log\dfrac{11}{15} + log \dfrac{35}{9} + log\dfrac{27}{77} = 0 \\ \\ [/tex]
Thus,
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Answers & Comments
Answer:
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:log\dfrac{ {a}^{2} }{ab} + log\dfrac{ {b}^{2} }{bc} + log\dfrac{ {c}^{2} }{ca} = 0\qquad \: \\ \\& \qquad \:\sf \:log_{2}( \sqrt{6} ) + log_{2}\left(\sqrt{\dfrac{2}{3} } \right) = 1\\ \\& \qquad \:\sf \:log\dfrac{11}{15} + log \dfrac{35}{9} + log\dfrac{27}{77} = 0 \end{aligned}} \qquad \: \\ \\ [/tex]
Step-by-step explanation:
[tex]\large\underline{\sf{Solution-1}}[/tex]
Given expression is
[tex]\sf \: log\dfrac{ {a}^{2} }{ab} + log\dfrac{ {b}^{2} }{bc} + log\dfrac{ {c}^{2} }{ca} \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: log\dfrac{ {a}}{b} + log\dfrac{ {b}}{c} + log\dfrac{ {c}}{a} \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{ {a}}{b} \times \dfrac{ {b}}{c} \times \dfrac{ {c}}{a}\right) \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: logx + logy = log(xy) \: }} \\ \\ [/tex]
[tex]\sf \: = \: log(1) \\ \\ [/tex]
[tex]\sf \: = \: 0 \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: log\dfrac{ {a}^{2} }{ab} + log\dfrac{ {b}^{2} }{bc} + log\dfrac{ {c}^{2} }{ca} = 0 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex]\large\underline{\sf{Solution-2}}[/tex]
Given expression is
[tex]\sf \: log_{2}( \sqrt{6} ) + log_{2}\left(\sqrt{\dfrac{2}{3} }\right) \\ \\ [/tex]
[tex]\sf \: = \: log_{2}\left(\sqrt{6} \times \sqrt{\dfrac{2}{3} } \right) \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: logx + logy = log(xy) \: }} \\ \\ [/tex]
[tex]\sf \: = \: log_{2}\left(\sqrt{2} \times \sqrt{2}\right) \\ \\ [/tex]
[tex]\sf \: = \: log_{2}(2) \\ \\ [/tex]
[tex]\sf \: = \: 1 \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: log_{2}( \sqrt{6} ) + log_{2}\left(\sqrt{\dfrac{2}{3} } \right) = 1 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex]\large\underline{\sf{Solution-3}}[/tex]
[tex]\sf \: log\dfrac{11}{15} + log \dfrac{35}{9} + log\dfrac{27}{77} \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{11}{15} \times \dfrac{35}{9} \times \dfrac{27}{77}\right) \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: logx + logy = log(xy) \: }} \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{1}{15} \times \dfrac{35}{9} \times \dfrac{27}{7}\right) \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{1}{15} \times \dfrac{35}{1} \times \dfrac{3}{7}\right) \\ \\ [/tex]
[tex]\sf \: = \: log\left(\dfrac{1}{5} \times \dfrac{35}{1} \times \dfrac{1}{7}\right) \\ \\ [/tex]
[tex]\sf \: = \: log\left(1\right) \\ \\ [/tex]
[tex]\sf \: = \: 0 \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: log\dfrac{11}{15} + log \dfrac{35}{9} + log\dfrac{27}{77} = 0 \\ \\ [/tex]
Thus,
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:log\dfrac{ {a}^{2} }{ab} + log\dfrac{ {b}^{2} }{bc} + log\dfrac{ {c}^{2} }{ca} = 0\qquad \: \\ \\& \qquad \:\sf \:log_{2}( \sqrt{6} ) + log_{2}\left(\sqrt{\dfrac{2}{3} } \right) = 1\\ \\& \qquad \:\sf \:log\dfrac{11}{15} + log \dfrac{35}{9} + log\dfrac{27}{77} = 0 \end{aligned}} \qquad \: \\ \\ [/tex]