Question

What is the natural log of $-0.9$?

Answer 1

Hint: First of all understand the term ‘natural logarithm’ and the value of base related to it. Now, use the fact that the log functions aren’t defined for negative arguments and use the properties of complex logarithms to get the answer. Write the given function as $\ln \left( -0.9 \right)=\ln \left( -1 \right)+\ln \left( 0.9 \right)$ by using the property $\log \left( mn \right)=\log m+\log n$. Now, the value of $\ln \left( 0.9 \right)$ can easily be found using the log table. To calculate the value of $\ln \left( -1 \right)$ write it as $\ln \left( {{i}^{2}} \right)$ where $i$ is the imaginary number $\sqrt{-1}$. Now, use the property of logarithm given as $\log \left( {{a}^{m}} \right)=m\log \left( a \right)$ to write $\ln \left( {{i}^{2}} \right)=2\ln \left( i \right)$. Use Euler’s formula in complex numbers given as ${{e}^{ix}}=\cos x+i\sin x$ and substitute the value of x equal to $\dfrac{\pi }{2}$ to substitute the value of $i$ in terms of e and simplify the value of $\ln \left( -1 \right)$.

Complete step by step solution:
Here we have been asked to find the value of natural log of $-0.9$. First let us understand the term natural log and its domain values.
In mathematics, natural log is a log function with the base value as the Euler’s number e = 2.71. It is denoted as $\ln \left( x \right)$. Now, for a log function to be defined its argument x must be greater than 0. In case x is negative we will not get any real value. Here in the above question we have to calculate the value of $\ln \left( -0.9 \right)$. Clearly we can see that the argument is negative so will not get any real value. Therefore, we have to move towards the concept of complex logarithms to get the answer.
Using the property of log given as $\log \left( mn \right)=\log m+\log n$, the given expression can be written as: -
$\Rightarrow \ln \left( -0.9 \right)=\ln \left( -1 \right)+\ln \left( 0.9 \right)$
Using the log table or calculator to calculate the value of $\ln \left( 0.9 \right)$ we get,
$\Rightarrow \ln \left( -0.9 \right)=\ln \left( -1 \right)-0.10536$
Now, we can write $-1={{i}^{2}}$ where $i$ the imaginary number $\sqrt{-1}$. So we get,
$\Rightarrow \ln \left( -0.9 \right)=\ln \left( {{i}^{2}} \right)-0.10536$
Using the formula $\log \left( {{a}^{m}} \right)=m\log \left( a \right)$ we get,
$\Rightarrow \ln \left( -0.9 \right)=2\ln \left( i \right)-0.10536$
Now, let us understand some properties of the Euler’s number e. According to Euler’s formula in complex numbers we have the relation ${{e}^{ix}}=\cos x+i\sin x$. Substituting the value of x equal to $\dfrac{\pi }{2}$ we get,
$\begin{align}
  & \Rightarrow {{e}^{i\dfrac{\pi }{2}}}=\cos \left( \dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\pi }{2} \right) \\
 & \Rightarrow {{e}^{i\dfrac{\pi }{2}}}=0+i\times 1 \\
 & \Rightarrow {{e}^{i\dfrac{\pi }{2}}}=i \\
\end{align}$
Substituting the above obtained value of $i$ in terms of e in the log function we get,
$\Rightarrow \ln \left( -0.9 \right)=2\ln \left( {{e}^{i\dfrac{\pi }{2}}} \right)-0.10536$
Again using the formula $\log \left( {{a}^{m}} \right)=m\log \left( a \right)$ and the value $\ln e=1$ we get,
$\begin{align}
  & \Rightarrow \ln \left( -0.9 \right)=2\times i\dfrac{\pi }{2}-0.10536 \\
 & \Rightarrow \ln \left( -0.9 \right)=\pi i-0.10536 \\
 & \therefore \ln \left( -0.9 \right)=-0.10536+\pi i \\
\end{align}$
Hence, the above obtained complex number is our answer.

Note: Note that in mathematics we come across two types of log namely: - common logarithm and natural logarithm. The base values of both the log functions are fixed. For common log it is 10 and for natural log it is e. You can also use the formula ${{\log }_{a}}\left( m\div n \right)={{\log }_{a}}m-{{\log }_{a}}n$ to write $\ln \left( 0.9 \right)=2\ln 3-\left[ \ln 2+\ln 5 \right]$ which can be further simplified using the conversion $\ln x=2.303{{\log }_{10}}x$. From here either you need to remember the common log values of 2, 3 and 5 or use the log table to get the answer. Note that the complex number we have obtained as the answer is one of the solution of the given log function because we know that sine trigonometric functions are periodic in nature so the general solution would be $\ln \left( -0.9 \right)=-0.10536+\left( 2n+1 \right)\pi i$ where n is any integer.