Question

How do you evaluate ${{\log }_{3}}\left( \dfrac{1}{27} \right)$?

Answer 1

Hint: To solve this question, we need to use the properties of the logarithm function. From the properties of logarithm function, the base raised to some exponent is equal to the exponent, that is ${{\log }_{b}}{{\left( b \right)}^{m}}=m$. So we have to write the argument $\dfrac{1}{27}$ in the form of $3$ raised to some exponent. Applying the property ${{\log }_{b}}{{\left( b \right)}^{m}}=m$ on the obtained expression, we will get the final answer.

Complete step-by-step solution:
The expression given in the above question is ${{\log }_{3}}\left( \dfrac{1}{27} \right)$. Let us write it in the below equation as
$E={{\log }_{3}}\left( \dfrac{1}{27} \right)$
So we have a logarithm to a fraction. The base of the logarithm, indicated by the subscript to the logarithm function, is given to be equal to $3$. From the properties of the logarithm function, we know that the logarithm of the base raised to some exponent is equal to the exponent.
So for simplifying the above expression, we need to write the argument in the form of $3$ raised to some exponent. We know that twenty seven is equal to the cube of three. So we can write the above expression as
$E={{\log }_{3}}\left( \dfrac{1}{{{3}^{3}}} \right)$
Now, we know that one raised to any power is equal to one itself. So we can add the exponent of three to the numerator to get
\[\begin{align}
 & E={{\log }_{3}}\left( \dfrac{{{1}^{3}}}{{{3}^{3}}} \right) \\
 & \Rightarrow E={{\log }_{3}}{{\left( \dfrac{1}{3} \right)}^{3}} \\
\end{align}\]
Now, from the properties of the exponents we can write $\dfrac{1}{3}={{3}^{-1}}$ in the above expression to get
\[E={{\log }_{3}}{{\left( {{3}^{-1}} \right)}^{3}}\]
We know that a power raised to the base raised to some power is equal to the base raised to the product of the power, that is, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$. So the above expression can also be written as
\[E={{\log }_{3}}{{\left( 3 \right)}^{-3}}\]
From the properties of logarithm, we know that
${{\log }_{b}}{{\left( b \right)}^{m}}=m$
So finally the above expression becomes
\[E=-3\]

Hence, the given expression ${{\log }_{3}}\left( \dfrac{1}{27} \right)$ is simplified and is equal to \[-3\].

Note:
We can also use the property $\log \left( \dfrac{A}{B} \right)=\log A-\log B$ to solve the above question. By applying this property, we will obtain the expression as ${{\log }_{3}}\left( 1 \right)-{{\log }_{3}}\left( 27 \right)$. We know that the logarithm of one is equal to zero. So the expression will become \[-{{\log }_{3}}\left( 27 \right)\] which can be simplified by using the properties used in the above solution.