Question

What is $\ln ({i^2})$?

Answer 1

We will use the concepts of algebra related to exponents and logarithms. We will define logarithms and their properties and rules. Through these definitions, we will solve this problem. And we will also look at some standard results of logarithms too.

Complete answer:
Consider an equation ${a^x} = b$.
Here, $a$ is called the base and $b$ is called the result. But there is another term $x$ which is called the exponent. Exponent is the power to the base which defines the number of times the base is multiplied by itself.
Suppose, there is a term ${3^5}$, which means that, 3 is multiplied by itself for five times.
So, ${3^5} = 3 \times 3 \times 3 \times 3 \times 3$
In the equation ${a^x} = b$, the value of $a$ can be found by making it the subject of the equation.
$\Rightarrow a = {\left( b \right)^{\dfrac{1}{x}}}$ or $a = \sqrt[x]{b}$
But, to find the value of $x$, we need to apply a function called ‘Logarithm’.
So, it is defined as the reverse process of exponential function.
For equation ${a^x} = b$, logarithm can be applied as $x = {\log _a}b$
So, there are few rules or limits for a logarithm. They are,
(1) Logarithm is not defined for a negative value i.e., $\log n$ is not defined if $n < 0$.
(2) Logarithm of 1 to any base is always 0. $\Rightarrow {\log _a}1 = 0$, because ${a^0} = 1$ where $a \in R$
(3) Logarithm to base $e = 2.71$ is represented as $\ln b$ which is called natural logarithm.
So, now, in the question it is given as $\ln ({i^2})$.
And we all know that, $i$ is a complex which is defined as $i = \sqrt { - 1}$
So, $\ln ({i^2}) = \ln \left( {{{\left( {\sqrt { - 1} } \right)}^2}} \right)$
$\Rightarrow \ln ({i^2}) = \ln \left( { - 1} \right)$
But we all know that logarithm is not defined for a negative value.
So, we can conclude that, $\ln ({i^2})$ is not defined.

Note:
Actually, natural logarithm $\ln ({i^2})$ is not defined. This means that its value is imaginary. It has no definite value.
We can use some formulas in logarithm like ${\log _n}\left( {ab} \right) = {\log _n}a + {\log _n}b$
And ${\log _n}\left( {\dfrac{a}{b}} \right) = {\log _n}a - {\log _n}b$
Logarithm to base 10 is represented as just $\log a$ (without representing any base)