If log a-b-c-log b-c-a-log c-a-b then aa- bb- cc-A- 1B- 2C- -1D- None of these

The correct option is A 1Let each ratio be k and y = a^ab^bc^c Then log a=k(b c), log y=∑ a log a Now ∑ a log a=∑ ak(b c)=0 ∴ log y=0⇒ y=1 or a^ab^bc^c=1 ...

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Byju's Answer

Standard XII

Mathematics

Fundamental Laws of Logarithms

If log a/b-c=...

Question

If $\frac{l o g a}{b - c} = \frac{l o g b}{c - a} = \frac{l o g c}{a - b}$ then $a^{a} . b^{b} . c^{c} =$

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-1

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None of these

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Solution

The correct option is A 1
Let each ratio be k and $y = a^{a} b^{b} c^{c}$
Then $l o g a = k (b - c), l o g y = \sum a l o g a$
Now $\sum a l o g a = \sum a k (b - c) = 0$
$∴ l o g y = 0 \Rightarrow y = 1$
or $a^{a} b^{b} c^{c} = 1$

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If log ab−c=log bc−a=log ca−b then aa.bb.cc=

The correct option is A 1Let each ratio be k and y=aabbcc Then log a=k(b−c),log y=∑a log a Now ∑a log a=∑ak(b−c)=0 ∴ log y=0⇒y=1 or aabbcc=1

If $\frac{l o g a}{b - c} = \frac{l o g b}{c - a} = \frac{l o g c}{a - b}$ then $a^{a} . b^{b} . c^{c} =$

The correct option is A 1
Let each ratio be k and $y = a^{a} b^{b} c^{c}$
Then $l o g a = k (b - c), l o g y = \sum a l o g a$
Now $\sum a l o g a = \sum a k (b - c) = 0$
$∴ l o g y = 0 \Rightarrow y = 1$
or $a^{a} b^{b} c^{c} = 1$