Basic of Logarithms
easy

$7\log \left( {{{16} \over {15}}} \right) + 5\log \left( {{{25} \over {24}}} \right) + 3\log \left( {{{81} \over {80}}} \right)= . . . .$

A

$0$

B

$1$

C

$\log 2$

D

$\log 3$

Solution

(c) Given expression = $\log \left( {{{{{16}^7}} \over {{{15}^7}}}.{{{{25}^5}} \over {{{24}^5}}}.{{{{81}^3}} \over {{{80}^3}}}} \right) = \log 2$.

Standard 11
Mathematics

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