The value of log-b-a-log-c-b-log-a-c -is0-log -abc-log-a-a-1-

The correct options are C log_a a D 1 log_b a.log_c b.log_a c =log a/log b.log b/log c. log c/log a Using identity, log_b a=log_m a/log_m b =log_a a = ...

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Byju's Answer

Standard X

Mathematics

Properties of Logarithms

The value of ...

Question

The value of $l o g_{b} a . l o g_{c} b . l o g_{a} c$ is

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$l o g (a b c)$

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$l o g_{a} a$

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Solution

The correct options are
C
$l o g_{a} a$

D
1

$l o g_{b} a . l o g_{c} b . l o g_{a} c$

$= \frac{l o g a}{l o g b} . \frac{l o g b}{l o g c} . \frac{l o g c}{l o g a}$ Using identity, $l o g_{b} a = \frac{l o g_{m} a}{l o g_{m} b}$

$= l o g_{a} a$

$= 1$ $∵ l o g_{a} a = 1$

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The value of logb a.logc b.loga c is

The correct options are C loga a D 1 logb a.logc b.loga c =log alog b.log blog c.log clog a Using identity, logb a=logm alogm b =loga a =1 ∵loga a=1

The value of $l o g_{b} a . l o g_{c} b . l o g_{a} c$ is

The correct options are
C
$l o g_{a} a$

D
1

$l o g_{b} a . l o g_{c} b . l o g_{a} c$

$= \frac{l o g a}{l o g b} . \frac{l o g b}{l o g c} . \frac{l o g c}{l o g a}$ Using identity, $l o g_{b} a = \frac{l o g_{m} a}{l o g_{m} b}$

$= l o g_{a} a$

$= 1$ $∵ l o g_{a} a = 1$