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Logarithms

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Question : 53 of 85

Marks: +1, -0

If

\log_{a} bc = x, \log_{b} ca = y, \log_{c} ab = z,

then the value of

\;\frac{1}{x+1}+\;\frac{1}{y+1}+\;\frac{1}{z+1}

0
1
$-1$
none

Solution:

Concept:

Use the property

\log_a m + \log_a n = \log_a (mn)

and change of base to express each term as

\log_{abc} a

, etc.

Explanation:

Given

x = \log_a (bc)

.

Then

x+1 = \log_a (bc) + 1 = \log_a (bc) + \log_a a = \log_a (abc)

.

Similarly,

y+1 = \log_b (ca) + 1 = \log_b (abc)

.

And

z+1 = \log_c (ab) + 1 = \log_c (abc)

.

Now,

\frac{1}{x+1} = \frac{1}{\log_a (abc)} = \log_{abc} a

.

Similarly,

\frac{1}{y+1} = \log_{abc} b

,

\frac{1}{z+1} = \log_{abc} c

.

Thus the sum is:

\log_{abc} a + \log_{abc} b + \log_{abc} c = \log_{abc} (abc) = 1

.

Answer:

1

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