Question

The value of ${{\log }_{10}}2+16{{\log }_{10}}\left( \dfrac{16}{15} \right)+12{{\log }_{10}}\left( \dfrac{25}{24} \right)+7{{\log }_{10}}\left( \dfrac{81}{80} \right)$ is:
(a) 3
(b) 2
(c) 1
(d) 0

Answer 1

Hint: Use the formula given by: \[{{\log }_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n\] to simplify the terms. Now, write the argument of each term as the product of their prime factors and use two different identities of logarithm given by: ${{\log }_{a}}\left( m\times n \right)={{\log }_{a}}m+{{\log }_{a}}n$ and \[{{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m\] to simplify the terms further. Cancel all the common terms and use the formula: ${{\log }_{a}}a=1$ to get the final answer.

Complete step-by-step answer:
Let us assume the value of the given expression as ‘E’. Therefore,
$E={{\log }_{10}}2+16{{\log }_{10}}\left( \dfrac{16}{15} \right)+12{{\log }_{10}}\left( \dfrac{25}{24} \right)+7{{\log }_{10}}\left( \dfrac{81}{80} \right)$
Using the formula: \[{{\log }_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n\], we get,
$E={{\log }_{10}}2+16{{\log }_{10}}16-16{{\log }_{10}}15+12{{\log }_{10}}25-12{{\log }_{10}}24+7{{\log }_{10}}81-7{{\log }_{10}}80$
Writing the arguments of the logarithm as the products of their primes, we get,
\[\begin{align}
  & E={{\log }_{10}}2+16{{\log }_{10}}\left( 2\times 2\times 2\times 2 \right)-16{{\log }_{10}}\left( 3\times 5 \right)+12{{\log }_{10}}\left( {{5}^{2}} \right)-12{{\log }_{10}}\left( 3\times 2\times 2\times 2 \right) \\
 & \text{ }+7{{\log }_{10}}\left( 3\times 3\times 3\times 3 \right)-7{{\log }_{10}}\left( 2\times 2\times 2\times 2\times 5 \right) \\
 & \Rightarrow E={{\log }_{10}}2+16{{\log }_{10}}\left( {{2}^{4}} \right)-16{{\log }_{10}}\left( 3\times 5 \right)+12{{\log }_{10}}\left( {{5}^{2}} \right)-12{{\log }_{10}}\left( 3\times {{2}^{3}} \right)+7{{\log }_{10}}\left( {{3}^{4}} \right)-7{{\log }_{10}}\left( {{2}^{4}}\times 5 \right) \\
\end{align}\]
Now, using the identity: ${{\log }_{a}}\left( m\times n \right)={{\log }_{a}}m+{{\log }_{a}}n$ and \[{{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m\], we get,
\[\begin{align}
  & E={{\log }_{10}}2+64{{\log }_{10}}2-16{{\log }_{10}}3-16{{\log }_{10}}5+24{{\log }_{10}}5-12{{\log }_{10}}3-36{{\log }_{10}}2+28{{\log }_{10}}3-28{{\log }_{10}}2-7{{\log }_{10}}5 \\
 & \Rightarrow E=65{{\log }_{10}}2-64{{\log }_{10}}2-28{{\log }_{10}}3+28{{\log }_{10}}3+24{{\log }_{10}}5-23{{\log }_{10}}5 \\
 & \Rightarrow E={{\log }_{10}}2+{{\log }_{10}}5 \\
\end{align}\]
Now, using the identity: ${{\log }_{a}}\left( m\times n \right)={{\log }_{a}}m+{{\log }_{a}}n$, we get,
$\begin{align}
  & E={{\log }_{10}}\left( 2\times 5 \right) \\
 & \Rightarrow E={{\log }_{10}}\left( 10 \right) \\
\end{align}$
Using the formula: ${{\log }_{a}}a=1$, we have,
$E=1$
Hence, option (c) is the correct answer.

Note: One may note that properties of logarithms used in the above solution are very helpful in simplifying the problem. We do not have to simplify the above expression by calculating the divisions given in the argument of logarithms. It will be a lengthy process and we will need the help of a log table for calculations, which is not available to us. It is also necessary to write the given arguments as the product of primes so that we can use the formula: ${{\log }_{a}}\left( m\times n \right)={{\log }_{a}}m+{{\log }_{a}}n$ for further simplification.