Question

Using the identity \[{{a}^{{{\log }_{a}}n}},\] find ${{3}^{\dfrac{-1}{2}{{\log }_{3}}9}}$ ,
(a) 0.33
(b) – 0.33
(c) 0.66
(d) – 0.66

Answer 1

Hint: To solve this question, we have to convert ${{3}^{\dfrac{-1}{2}{{\log }_{3}}9}}$ into the form \[{{a}^{{{\log }_{a}}n}}\] , but the factor $\dfrac{-1}{2}$ in the power is the reason we can't express this in the required form. To remove $\left( \dfrac{-1}{2} \right)$ , we will use a logarithmic property which is given as:
$a{{\log }_{x}}y={{\log }_{x}}{{y}^{a}}$

Complete step-by-step answer:
The question demands that we have to find the value of ${{3}^{\dfrac{-1}{2}{{\log }_{3}}9}}$ only with the help of the identity given in question. So, we will have to convert ${{3}^{\dfrac{-1}{2}{{\log }_{3}}9}}$ into the required form. We know that the factor $\left( \dfrac{-1}{2} \right)$ is extra. So, we will have to remove this. To remove this, we are going to use a logarithmic identity which is as follows:
$a{{\log }_{x}}y={{\log }_{x}}{{y}^{a}}$
So, using this identity we can write the power term i.e. $\dfrac{-1}{2}{{\log }_{3}}9$ as shown below
$\dfrac{-1}{2}{{\log }_{3}}9={{\log }_{3}}{{9}^{\dfrac{-1}{2}}}...............\left( i \right)$
Rearranging the equation (i) we will get following:
$\Rightarrow \dfrac{-1}{2}{{\log }_{3}}9={{\log }_{3}}{{\left( \dfrac{1}{9} \right)}^{\dfrac{1}{2}}}................\left( ii \right)$
Now, using the exponential identity, as shown below
$\dfrac{{{a}^{n}}}{{{b}^{n}}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}$
Using the above identity in equation (ii), we will get following result
$\begin{align}
  & \Rightarrow \dfrac{-1}{2}{{\log }_{3}}9={{\log }_{3}}\dfrac{{{\left( 1 \right)}^{\dfrac{1}{2}}}}{{{\left( 9 \right)}^{\dfrac{1}{2}}}} \\
 & \Rightarrow \dfrac{-1}{2}{{\log }_{3}}9={{\log }_{3}}\left( \dfrac{1}{3} \right)................\left( iii \right) \\
\end{align}$
Now, we will put the value of $\dfrac{-1}{2}{{\log }_{3}}9$ in the question given. After doing this, we will get the following result:
${{3}^{{{\log }_{3}}\left( \dfrac{1}{3} \right)}}$
Now the above term represents the left-hand side of the identity. So, in our case, the value of a becomes 3 and value of n becomes $\left( \dfrac{1}{3} \right)$ .
Thus, as we know that, according to the identity, we have
$\Rightarrow {{3}^{\log \left( \dfrac{1}{3} \right)}}=\left( \dfrac{1}{3} \right)$
Therefore, we can say that,
${{3}^{\dfrac{-1}{2}{{\log }_{3}}9}}=\left( \dfrac{1}{3} \right)$
The fractional form of $\dfrac{1}{3}$ is 0.33.
Hence, option (a) is correct.

Note: The identity given in question is not valid everywhere. Here, to make the identity valid, the value of n > 0. Also, the value of a should be positive, it should not be equal to 1. Also, we can eliminate the options (b) and (d) directly as the value of exponent cannot be negative.