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Logarithms

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Question : 55 of 85

Marks: +1, -0

If

a=1+\log_{x} yz, b=1+\log_{y} zx, c=1+\log_{z} xy,

ab+bc+ca=-

0
$2abc$
$a^{2}+b^{2}+c^{2}$
abc.

Solution:

Concept:

Use the identity

1/a + 1/b + 1/c = 1

derived from log properties.

Explanation:

Rewrite each term:

a = 1 + \log_x(yz) = \log_x(xyz)

,

b = \log_y(xyz)

,

c = \log_z(xyz)

.

Then

1/a = \log_{xyz}(x)

,

1/b = \log_{xyz}(y)

,

1/c = \log_{xyz}(z)

.

Since

\log_{xyz}(x) + \log_{xyz}(y) + \log_{xyz}(z) = \log_{xyz}(xyz) = 1

, we have

1/a + 1/b + 1/c = 1

.

Multiply both sides by

abc

:

bc + ac + ab = abc

.

Answer:

abc

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