Question

The value of ${{\log }_{2}}\dfrac{1}{8}$ is
(a) 2
(b) 0
(c) -1
(d) -3

Answer 1

Hint: To solve this problem, we have to use the logarithmic identity of power. According to this identity, ${{\log }_{a}}{{\left( b \right)}^{m}}=m{{\log }_{a}}b$ . This will help us to simplify the problem to a large extent. We will also have to use the basic identity of logarithms, that says, ${{\log }_{a}}a=1$ . Using these identities, we can easily get the required value.

Complete step by step solution:
According to the definition of logarithms, if we have $a>0$ and b as a positive number not equal to 1, then ${{\log }_{b}}a$ is called the logarithm of a with base b.
We also know that a logarithm is an inverse function of exponential.
For example, ${{2}^{5}}=32$ is same as ${{\log }_{2}}32=5$ .
In our problem, we have to find the value of ${{\log }_{2}}\dfrac{1}{8}$ .
So, let us assume a variable $x$ such that $x={{\log }_{2}}\dfrac{1}{8}$ .
We can also write the above equation as
$x={{\log }_{2}}{{\left( 8 \right)}^{-1}}...\left( i \right)$
We know the logarithmic identity ${{\log }_{a}}{{\left( b \right)}^{m}}=m{{\log }_{a}}b$ .
So, we can use this identity in equation (i) to simplify it as follows,
$x=-{{\log }_{2}}8...\left( ii \right)$
We also know that ${{2}^{3}}=8$ and so we can substitute 8 in equation (ii),
 $x=-{{\log }_{2}}{{2}^{3}}...\left( iii \right)$
We can once again use the logarithmic identity ${{\log }_{a}}{{\left( b \right)}^{m}}=m{{\log }_{a}}b$ .
Hence, equation (iii) will be reduced to
$x=-3{{\log }_{2}}2$
We all are very well aware of the fact that ${{\log }_{a}}a=1$ .
Thus, we have
$x=-3$.

So, the correct answer is “Option (d)”.

Note: We must know that when the base of any logarithm is 2, then the logarithm is called binary logarithm. We can also the above problem using the identity ${{\log }_{a}}m-{{\log }_{a}}n={{\log }_{a}}\left( \dfrac{m}{n} \right)$ , and the fact that logarithm of one for any base is zero.