If x-log ab c- y-log bc a- z-log ca b- then which of the following is equal to 1A- x-y-z B- xyzC- None of theseD- 1-x-1-1-y-1-1-z-1

The correct option is D (1+x)^ 1+(1+y)^ 1+(1+z)^ 1x=log_a bc ⇒ 1+x=log_a a+ log_a bc = log_a abc ∴ (1+x)^ 1=log_abca ∴ (1+x)^ 1+(1+y)^ 1+(1+z)^ 1=log_abc a+log ...

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Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

If x=log ab c...

Question

If $x = l o g_{a} (b c), y = l o g_{b} (c a), z = l o g_{c} (a b),$ then which of the following is equal to 1

x + y + z

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(1 + x)^{- 1} + (1 + y)^{- 1} + (1 + z)^{- 1}

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xyz

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None of these

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Solution

The correct option is B $(1 + x)^{- 1} + (1 + y)^{- 1} + (1 + z)^{- 1}$
$x = l o g_{a} b c \Rightarrow 1 + x = l o g_{a} a + l o g_{a} b c = l o g_{a} a b c$
$∴ (1 + x)^{- 1} = l o g_{a b c} a$
$∴ (1 + x)^{- 1} + (1 + y)^{- 1} + (1 + z)^{- 1} = l o g_{a b c} a + l o g_{a b c} c + l o g_{a b c} b$
$= l o g_{a b c} a b c = 1$

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If x=loga(bc), y=logb(ca), z=logc(ab), then which of the following is equal to 1

The correct option is B (1+x)−1+(1+y)−1+(1+z)−1x=logabc⇒1+x=logaa+logabc=logaabc ∴(1+x)−1=logabca ∴ (1+x)−1+(1+y)−1+(1+z)−1=logabc a+logabc c+logabc b =logabcabc=1

If $x = l o g_{a} (b c), y = l o g_{b} (c a), z = l o g_{c} (a b),$ then which of the following is equal to 1