Question

How do you evaluate ${\log _2}\left( {\dfrac{1}{2}} \right)$ ?

Answer 1

Hint: In this question they have given a logarithmic function to evaluate. From the definition of the logarithm we have that, if $x > 0$ and $b$ is a positive constant number and $b$ is not equal to one, then we have ${\log _b}x = y$ which is equivalent to $x = {b^y}$ . By using this definition of logarithmic function we can arrive at the solution.

Complete step by step answer:
Here in this question, they have given a logarithmic function which is ${\log _2}\left( {\dfrac{1}{2}} \right)$to evaluate. The function ${\log _2}\left( {\dfrac{1}{2}} \right)$ is read as log $\dfrac{1}{2}$ to the base $2$ . Now we need to find the value of the log $\dfrac{1}{2}$ to the base $2$.
From the definition of logarithm we have that, if $x > 0$ and $b$ is a positive constant number and $b$ is not equal to one, then we have ${\log _b}x = y$ which is equivalent to $x = {b^y}$.
By comparing the given function which is ${\log _2}\left( {\dfrac{1}{2}} \right)$ with the definition of logarithmic function ${\log _b}x = y$ , we have $x = \dfrac{1}{2}$ and$b = 2$, $y = ?$ So here we need to find the value of $y$. to find the value of $y$ we can make use of $x = {b^y}$ by the definition of logarithm.
Therefore on substituting all the values in the expression $x = {b^y}$ to find $y$, we get
$\dfrac{1}{2} = {2^y}$
In order to find the value of $y$ which is an exponent term in the above expression, first, we need to keep the base values the same. Then only we can compare the exponent terms. So now to make the base terms as same, we can rewrite the above expression as below:
$ \Rightarrow {2^{ - 1}} = {2^y}$ (when we write denominator terms in the numerator it will become inverse value)
As we can see that the bases are the same in the above equation which is $2$, we can equate the exponent terms. Therefore, we get
$ \Rightarrow - 1 = y$ Or $y = - 1$.

Therefore, the value of log $\dfrac{1}{2}$ to the base $2$ is $ - 1$ i.e., ${\log _2}\left( {\dfrac{1}{2}} \right) = - 1$.

Note:
In this type of problems you can directly make use of the definition of the logarithmic function, one thing you need to remember is while simplifying for the common base make sure you get the same base otherwise you cannot equate the exponent terms, only if the base is same you can equate the exponents.