Question

Simplify the expression $\ln \left( {\dfrac{1}{{{e^2}}}} \right)$.

Answer 1

Hint: $\;\ln \left( {{P^q}} \right) = q \times \ln \left( P \right)$- Power Rule
By using power rules and basic logarithmic identities we can simplify the above question.
Also we know $\ln \left( {{e^y}} \right) = y$ since the exponential and logarithmic functions are inverse in nature.
So by using the above two identities we would be able to simplify the expression$\ln \left(
{\dfrac{1}{{{e^2}}}} \right)$. So by using the logarithmic expressions and identities we can simplify$\ln
\left( {\dfrac{1}{{{e^2}}}} \right)$.

Complete step by step solution:
Given
$\ln \left( {\dfrac{1}{{{e^2}}}} \right).............................\left( i \right)$
Now we have to simplify the logarithmic term and the term $\left( {\dfrac{1}{{{e^2}}}} \right)$along
with it.
We can write $\left( {\dfrac{1}{{{e^2}}}} \right)$ as ${e^{ - 2}}$i.e. $\left( {\dfrac{1}{{{e^2}}}} \right) =
{e^{ - 2}}$in accordance with the negative exponential rule.

So from (i) we can write
$\ln \left( {\dfrac{1}{{{e^2}}}} \right) = \ln \left( {{e^{ - 2}}} \right)..............................\left( {ii} \right)$
Also $\ln \left( {{e^y}} \right) = y.......................\left( {iii} \right)$ since the exponential and logarithmic functions are inverse in nature.
Now comparing (ii) in (iii) we get:
$\ln \left( {{e^{ - 2}}} \right) = - 2..................\left( {iv} \right)$
Therefore from (iv) we can write:
$\ln \left( {\dfrac{1}{{{e^2}}}} \right) = - 2$

Alternative method:
We know the Quotient Rule: $
\\
\;\ln \left( {\dfrac{P}{Q}} \right) = \ln \left( P \right) - \ln \left( Q \right) \\
\\
$
So on comparing $\ln \left( {\dfrac{1}{{{e^2}}}} \right)$ with the Quotient Rule we can observe that$P =
1\;and\;Q = {e^2}$.
Therefore by using the Quotient Rule and other logarithmic properties we can simply $\ln \left(
{\dfrac{1}{{{e^2}}}} \right)$ directly and simply.
So applying Quotient Rule in (i) we get:
$\ln \left( {\dfrac{1}{{{e^2}}}} \right) = \ln \left( 1 \right) - \ln \left( {{e^2}} \right).........................(v)$
Also we know $\ln \left( 1 \right) = 0$ which is a basic logarithmic identity is
And from (iii) $\ln \left( {{e^2}} \right) = 2$ since we know $\ln \left( {{e^y}} \right) = y$
On substituting these values in (v) we get:
$
\ln \left( {\dfrac{1}{{{e^2}}}} \right) = 0 - (2) \\
\;\;\;\;\;\;\;\;\;\;\; = - 2 \\
$

Therefore: $\ln \left( {\dfrac{1}{{{e^2}}}} \right) = - 2$which is our final answer.

Note:Some of the logarithmic properties useful for solving questions of derivatives are listed below:
$
1.\;\ln \left( {PQ} \right) = \ln \left( P \right) + \ln \left( Q \right) \\

2.\;\ln \left( {\dfrac{P}{Q}} \right) = \ln \left( P \right) - \ln \left( Q \right) \\

3.\;\ln \left( {{P^q}} \right) = q \times \ln \left( P \right) \\
$
Equations ‘1’ ‘2’ and ‘3’ are called Product Rule, Quotient Rule and Power Rule respectively.
Also another basic identities which are necessary to solve logarithmic questions are given below:
$\ln 1 = 0$where ‘a’ is any real number.