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Logarithms

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Question : 67 of 85

Marks: +1, -0

\log \frac{75}{16}-2\log \frac{5}{9}+\log \frac{32}{243}

reduces to-

$2\log 2$
$5\log 2$
$\log 2$
$4\log 2$

Solution:

Concept:

Use logarithm properties:

\log a - \log b = \log \frac{a}{b}

and

n\log a = \log a^n

.

Explanation:

Combine terms step by step:

\log \frac{75}{16} - 2\log \frac{5}{9} + \log \frac{32}{243}

= \log \frac{75}{16} - \log \left(\frac{5}{9}\right)^2 + \log \frac{32}{243}

= \log \frac{75}{16} - \log \frac{25}{81} + \log \frac{32}{243}

Combine first two logs:

\log \left( \frac{75}{16} \div \frac{25}{81} \right) = \log \left( \frac{75}{16} \times \frac{81}{25} \right) = \log \frac{75 \times 81}{16 \times 25} = \log \frac{3 \times 81}{16} = \log \frac{243}{16}

Now add the third log:

\log \frac{243}{16} + \log \frac{32}{243} = \log \left( \frac{243}{16} \times \frac{32}{243} \right) = \log \frac{32}{16} = \log 2

Answer:

\log 2

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