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Logarithms

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Question : 73 of 85

Marks: +1, -0

Find the value of

lm+m n+nl-lm n

l=1+\log_{a} bc, m=1+\log_{h} ca, n=1+\log_{c} ab

$0$
$1$
$-1$
$3$

Solution:

Concept:

Use logarithm properties to rewrite

l

,

m

,

n

as

\log_a(abc)

,

\log_b(abc)

,

\log_c(abc)

respectively.

Explanation:

l = 1 + \log_a(bc) = \log_a a + \log_a b + \log_a c = \log_a(abc)

.

Similarly,

m = \log_b(abc)

and

n = \log_c(abc)

.

Let

k = abc

. Then

l = \frac{\ln k}{\ln a}

,

m = \frac{\ln k}{\ln b}

,

n = \frac{\ln k}{\ln c}

.

Compute

lm + mn + nl - lmn

:

= \frac{(\ln k)^2}{\ln a \ln b} + \frac{(\ln k)^2}{\ln b \ln c} + \frac{(\ln k)^2}{\ln c \ln a} - \frac{(\ln k)^3}{\ln a \ln b \ln c}

= \frac{(\ln k)^2}{\ln a \ln b \ln c}\left( \ln c + \ln a + \ln b - \ln k \right)

= \frac{(\ln k)^2}{\ln a \ln b \ln c}\left( \ln(abc) - \ln k \right) = 0

.

Answer:

0

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