Question

Evaluate ${\log _3}\left( {\dfrac{1}{{81}}} \right)$.

Answer 1

$\\\ \;{\log _b}\left( {\dfrac{P}{Q}} \right) = {\log _b}\left( P \right) - {\log _b}\left( Q \right) \\\ \\\$- Quotient Rule
$\;{\log _b}\left( {{P^q}} \right) = q \times {\log _b}\left( P \right)$- Power Rule
So by using the Quotient Rule and Power Rule we can simplify and thus solve the above given
logarithmic term. The logarithmic term has to be simplified using the logarithmic laws and thus the value has to be found.

Complete step by step solution:
Given
${\log _3}\left( {\dfrac{1}{{81}}} \right)..............................\left( i \right)$
Now by using the Quotient Rule $\\\ \;{\log _b}\left( {\dfrac{P}{Q}} \right) = {\log _b}\left( P \right) - {\log _b}\left( Q \right) \\\ \\\$
Converting (i) in terms of Quotient rule we get:
${\log _3}\left( {\dfrac{1}{{81}}} \right) = {\log _3}\left( 1 \right) - {\log _3}\left( {81} \right)............................\left( {ii} \right)$
Now on analyzing (ii) we know that ${\log _3}\left( 1 \right) = 0......................(iii)$
Also we can write $81 = {3^4}......................\left( {iv} \right)$
So we can substitute (iv) and (v) in (ii):
We get ${\log _3}\left( {\dfrac{1}{{81}}} \right) = 0 - {\log _3}\left( {{3^4}} \right)$
$\Rightarrow {\log _3}\left( {\dfrac{1}{{81}}} \right) = - {\log _3}\left( {{3^4}} \right)............................\left( v \right)$
Now substituting Power Rule $\;{\log _b}\left( {{P^q}} \right) = q \times {\log _b}\left( P \right)$in (vi):
We get ${\log _3}\left( {\dfrac{1}{{81}}} \right) = - 4{\log _3}\left( 3 \right)......................\left( {vi} \right)$

Now analyzing (vii) we know ${\log _3}\left( 3 \right) = 1$
Substituting this value in (vii) we get ${\log _3}\left( {\dfrac{1}{{81}}} \right) = - 4...........................\left( {vii} \right)$
Therefore the answer on evaluating ${\log _3}\left( {\dfrac{1}{{81}}} \right)$ is$- 4$.

Alternative Method:
Given
${\log _3}\left( {\dfrac{1}{{81}}} \right)$
Let ${\log _3}\left( {\dfrac{1}{{81}}} \right) = q.....................\left( {viii} \right)$
Also if ${\log _b}\left( y \right) = q$ $\Rightarrow {b^q} = y............................\left( {ix} \right)$

(ix) Represents the exponential form of a logarithmic function.
Here $b = 3$ and also $81 = {3^4}$ such that we can write $\left( {\dfrac{1}{{81}}} \right) = {3^{ - 4}}$
$\Rightarrow {3^{ - 4}} = y$
Now substituting the values in (ix) such that:
${3^q} = {3^{ - 4}}...................\left( x \right)$ Here the base is the same on both sides which is 3, so we can directly compare and write the answer directly.
Therefore on comparing we get $q = - 4$.
Now from (viii) ${\log _3}\left( {\dfrac{1}{{81}}} \right) = q$
Hence${\log _3}\left( {\dfrac{1}{{81}}} \right) = - 4$.
This method is comparatively easier and direct in comparison to the 1 st method.

Note: Some of the logarithmic properties useful for solving questions of derivatives are listed below:
$1.\;{\log _b}\left( {PQ} \right) = {\log _b}\left( P \right) + {\log _b}\left( Q \right) \\\ 2.\;{\log _b}\left( {\dfrac{P}{Q}} \right) = {\log _b}\left( P \right) - {\log _b}\left( Q \right) \\\ 3.\;{\log _b}\left( {{P^q}} \right) = q \times {\log _b}\left( P \right) \\\$
Equations ‘1’ ‘2’ and ‘3’ are called Product Rule, Quotient Rule and Power Rule respectively.
Also two basic identities which are necessary to solve logarithmic questions are given below:
${\log _a}1 = 0$ where ‘a’ is any real number.
${\log _a}a = 1$ where ‘a’ is any real number.