Question

The value of $7\log \left( \dfrac{16}{15} \right)+5\log \left( \dfrac{25}{24} \right)+3\log \left( \dfrac{81}{80} \right)$ is equal to:
(A) $\log 2$
(B) $\log 3$
(C) $\log 5$
(D) $\log 6$

Answer 1

Hint: For answering this question we need to simplify the expression that we have from the question to find the value of $7\log \left( \dfrac{16}{15} \right)+5\log \left( \dfrac{25}{24} \right)+3\log \left( \dfrac{81}{80} \right)$. By using the following logarithm identities we can simplify it. The identities are stated as:
$\log \left( \dfrac{a}{b} \right)=\log a-\log b$ and $\log \left( ab \right)=\log a+\log b$ and $\log \left( {{a}^{m}} \right)=m\log a$ .

Complete step by step answer:
Now from considering this question we have the expression as $7\log \left( \dfrac{16}{15} \right)+5\log \left( \dfrac{25}{24} \right)+3\log \left( \dfrac{81}{80} \right)$ .
By using the logarithm identity $\log \left( \dfrac{a}{b} \right)=\log a-\log b$ and simplifying this we will have it as
$7\log \left( 16 \right)-7\log \left( 15 \right)+5\log \left( 25 \right)-5\log \left( 24 \right)+3\log \left( 81 \right)-3\log \left( 80 \right)$.
This expression we can expand it as
$7\log \left( {{2}^{4}} \right)-7\log \left( 3\times 5 \right)+5\log \left( {{5}^{2}} \right)-5\log \left( 3\times {{2}^{3}} \right)+3\log \left( {{3}^{4}} \right)-3\log \left( {{2}^{4}}\times 5 \right)$ .
This expression can be further simplified using the logarithm identity $\log \left( ab \right)=\log a+\log b$ and after that we will have it as
$7\log \left( {{2}^{4}} \right)-7\log \left( 3 \right)-7\log \left( 5 \right)+5\log \left( {{5}^{2}} \right)-5\log \left( 3 \right)-5\log \left( {{2}^{3}} \right)+3\log \left( {{3}^{4}} \right)-3\log \left( {{2}^{4}} \right)-3\log \left( 5 \right)$
For further simplifying this expression we will use the logarithm identity $\log \left( {{a}^{m}} \right)=m\log a$ and after that we will have it as
$7\left( 4 \right)\log \left( 2 \right)-7\log \left( 3 \right)-7\log \left( 5 \right)+5\left( 2 \right)\log \left( 5 \right)-5\log \left( 3 \right)-5\left( 3 \right)\log \left( 2 \right)+3\left( 4 \right)\log \left( 3 \right)-3\left( 4 \right)\log \left( 2 \right)-3\log \left( 5 \right)$
By further simplifying this expression we will have it as $28\log \left( 2 \right)-7\log \left( 3 \right)-7\log \left( 5 \right)+10\log \left( 5 \right)-5\log \left( 3 \right)-15\log \left( 2 \right)+12\log \left( 3 \right)-12\log \left( 2 \right)-3\log \left( 5 \right)$.
After performing calculations and simplifying it we will have it as $\left( 28-15-12 \right)\log \left( 2 \right)+\left( 12-7-5 \right)\log \left( 3 \right)+\left( 10-7-3 \right)\log \left( 5 \right)$ .
Hence, after all the calculations and simplifications and using all the necessary logarithm identities we will come to a conclusion and have the value of the expression $7\log \left( \dfrac{16}{15} \right)+5\log \left( \dfrac{25}{24} \right)+3\log \left( \dfrac{81}{80} \right)$ as $\log 2$ .

So, the correct answer is “Option A”.

Note: While answering questions of this type we should be clear with the calculations and we should also remember the logarithm identities. While calculating if we had made a mistake and write it as $\left( 28-15-12 \right)\log \left( 2 \right)+\left( 12-7-5 \right)\log \left( 3 \right)+\left( 10-7-4 \right)\log \left( 5 \right)$ we will have the conclusion as the answer is $\log 2-\log 5$ which is completely wrong.