Question

What is the value of the following expression: $2{{\log }_{8}}2-\dfrac{1}{3}{{\log }_{3}}9$
(a) 0
(b) 1
(c) 2
(d) $\dfrac{1}{3}$

Answer 1

Hint: First, define the various properties of logarithms. Use the properties ${{\log }_{b}}\left( {{a}^{n}} \right)=n{{\log }_{b}}a$ and ${{\log }_{\left( {{b}^{n}} \right)}}a=\dfrac{1}{n}{{\log }_{b}}a$ to evaluate the first and second terms separately. Then subtract the two results to get the final answer.

Complete step-by-step answer:

In this question, we need to find the value of the following expression: $2{{\log }_{8}}2-\dfrac{1}{3}{{\log }_{3}}9$

For this question, we will need to use the various logarithmic rules. Let us first define these logarithmic rules.

A logarithm is the power to which a number must be raised in order to get some other number.

If we take log with base b of a number a which is raised to the power n, then the result is the product of n and the log with base b of a.

i.e. ${{\log }_{b}}\left( {{a}^{n}} \right)=n{{\log }_{b}}a$

Another property is that if we take log with base b which is raised to the power n of a number a, then the result is the product of the reciprocal of n and the log with base b of a.
i.e. ${{\log }_{\left( {{b}^{n}} \right)}}a=\dfrac{1}{n}{{\log }_{b}}a$

We will use these two properties on the given expression to find its value.

First, let us evaluate the first term.

 $2{{\log }_{8}}2=2{{\log }_{{{2}^{3}}}}2$

So, we will use the second property here. Using that, we will get the following:
$2{{\log }_{8}}2=2\times \dfrac{1}{3}{{\log }_{2}}2=\dfrac{2}{3}{{\log }_{2}}2$
We know that log of a number a with base a is equal to 1, i.e. ${{\log }_{a}}a=1$
So, $2{{\log }_{8}}2=\dfrac{2}{3}$

                                        …(1)
Now, we will evaluate the second term.

$\dfrac{1}{3}{{\log }_{3}}9=\dfrac{1}{3}{{\log }_{3}}{{3}^{2}}$

We will use the first property here. Using that, we will get the following:

$\dfrac{1}{3}{{\log }_{3}}9=\dfrac{1}{3}\times 2{{\log }_{3}}3=\dfrac{2}{3}{{\log }_{3}}3$

We know that log of a number a with base a is equal to 1, i.e. ${{\log }_{a}}a=1$

 $\dfrac{1}{3}{{\log }_{3}}9=\dfrac{2}{3}$ …(2)
Now, using equations (1) and (2), we will get the following:

\[2{{\log }_{8}}2-\dfrac{1}{3}{{\log }_{3}}9=\dfrac{2}{3}-\dfrac{2}{3}=0\]

So, the answer is 0.

Hence, option (a) is correct.

Note: In this question, it is very important to know what a logarithm is, what are its properties and how they are used. Students may get confused between the properties ${{\log }_{b}}\left( {{a}^{n}} \right)=n{{\log }_{b}}a$ and ${{\log }_{\left( {{b}^{n}} \right)}}a=\dfrac{1}{n}{{\log }_{b}}a$. So understand them properly before using them.