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Logarithms

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Question : 18 of 85

Marks: +1, -0

If

\log \left(\frac{m}{n}\right) + \log \left(\frac{n}{m}\right) = \log (m+n)

then

$m+n=0$
$m+n=1$
$\frac{m}{n}=1$
$m-n=1$

Solution:

Concept:

Use the product rule of logarithms:

\log a + \log b = \log(ab)

to combine the left-hand side.

Explanation:

Given:

\log\left(\frac{m}{n}\right) + \log\left(\frac{n}{m}\right) = \log(m+n)

.

Apply the product rule:

\log\left(\frac{m}{n} \cdot \frac{n}{m}\right) = \log(1) = 0

.

Thus,

\log(m+n) = 0

.

Convert to exponential form:

m+n = 10^0 = 1

.

Answer:

Option B:

m+n=1

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