Question

How do you find the exact value of \[\ln {\text{ }}\sqrt[4]{{{e^3}}}\] ?

Answer 1

Hint: In this question the basic idea is to use the logarithm property which is given by ${\log _a}{\text{ }}{{\text{b}}^m} = m{\log _a}{\text{ b}}$ . Here in this question the base of logarithm is e i.e. natural logarithm and the value of ${\log _e}{\text{ e}}$ is 1.

Complete step by step answer:
Let I= \[\ln {\text{ }}\sqrt[4]{{{e^3}}}\]
As (ln x) can also be written as $({\log _e}{\text{ x)}}$
As it is known that $\sqrt[x]{{{a^y}}} = {a^{\dfrac{y}{x}}}$
$ \Rightarrow I = {\log _e}{\text{ }}{{\text{e}}^{\dfrac{3}{4}}}$
$\because {\text{ }}{\log _a}{\text{ }}{{\text{b}}^m} = m{\log _a}{\text{ b}}$
On comparing,
$a = e$
$b = e$
$m = \dfrac{3}{4}$
$ \Rightarrow I = {\log _e}{\text{ }}{{\text{e}}^{\dfrac{3}{4}}}$
$ \Rightarrow I = \dfrac{3}{4}{\log _e}{\text{ e}}$
$\because {\text{ lo}}{{\text{g}}_e}{\text{ e = 1}}$
$\therefore $ $I = \dfrac{3}{4}$
On dividing,
$ \Rightarrow I = 0.75$
$\therefore {\text{ The exact value of }}\ln {\text{ }}\sqrt[4]{{{e^3}}}{\text{ is 0}}{\text{.75}}$

Note:
 If the logarithm comes in the form ${\log _a}\dfrac{m}{n}$ , then it can be substituted by ${\log _a}m - {\log _a}n$ and if the logarithm comes in the form $lo{g_a}mn$ then it can be substituted by ${\log _a}mn = {\log _a}m + {\log _a}n$ . The value of ${\log _a}a$ is always 1. The final answer can be left in fraction form also. The natural logarithm can be converted into logarithm to the base 10 by using the equation given by $\ln {\text{ x = 2}}{\text{.303lo}}{{\text{g}}_{10}}{\text{ x}}$ . The base ‘e’ is also called the Euler number and it is a constant whose value is 2.718.