Question

How do you find the exact value of $\ln \left( {\dfrac{1}{{\sqrt e }}} \right)$ ?

Answer 1

Hint: For solving this particular problem we will use ${\log _e}{a^b} = b{\log _e}a$ , exponent rule may be used if $a$ is greater than zero. For simplifying the equation , we will use a logarithm property that is \[{\log _e}e = \ln e\] logarithm with base $e$ is equivalent to the natural logarithm .

Formula used:
We used logarithm property i.e., ${\log _e}{a^b} = b{\log _e}a$ , exponent rule may be used if $a$ is greater than zero. For simplifying the equation , we will use a logarithm property that is \[{\log _e}e = \ln e\] logarithm with base $e$ is equivalent to the natural logarithm .
We also used the identity ${\log _x}x = 1$ , this will always give us one .

Complete step by step solution:
Given $\ln \left( {\dfrac{1}{{\sqrt e }}} \right)$
We have to find the exact value of the given expression.
We can write this as ,
$ \Rightarrow \ln \left( {\dfrac{1}{{\sqrt e }}} \right) = \ln \left( {{e^{ - \dfrac{1}{2}}}} \right)$
Now, applying the identity $\ln {a^b} = b\ln a$ , we will get the following ,
$ = - \dfrac{1}{2}\ln e$
We also know that \[{\log _e}e = \ln e\] logarithm with base $e$ is equivalent to the natural logarithm ,
Therefore, we can write ,
$ = - \dfrac{1}{2}{\log _e}e$
Now using the identity ${\log _x}x = 1$ , we will get the following result ,
$ = - \dfrac{1}{2}$

Therefore, we get the required result that is $ - \dfrac{1}{2}$

Note: The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero. While performing logarithm properties we have to remember certain conditions , our end result must satisfy the domain of that logarithm .