Answer
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Hint: Here we will proceed with the solution as we know the range of $\sin x = [1, - 1]$ which is required to solve this problem.
Here we need to find the range of given value that is ${\log _e}(\sin x)$
As we know that the domain of logarithmic functions are of positive value only
Since we know the range of $\sin x$ is $[ - 1,1]$
Then the domain value of above log function would be $(0,1]$
So, now to get the range of ${\log _e}(\sin x)$
Let us substitute $x = 0$ in the given function ${\log _e}(\sin x)$
i.e.
For $x \to 0 \Rightarrow {\log _e}x = - \infty $
Now let us substitute $x = 1$ in the given function ${\log _e}(\sin x)$
i.e.
For$x \to 1 \Rightarrow {\log _e}x = 0$
Hence from this we can say that the range of given function ${\log _e}(\sin x)$=$( - \infty ,0]$
NOTE: In this particular problem we know the range of $\sin x$ is $[ - 1,1]$ and domain of log function is $(0,1]$ so by substituting the domain values ($x$ values) in the given function i.e. ${\log _e}(\sin x)$ . We will get the range of the given function.
Here we need to find the range of given value that is ${\log _e}(\sin x)$
As we know that the domain of logarithmic functions are of positive value only
Since we know the range of $\sin x$ is $[ - 1,1]$
Then the domain value of above log function would be $(0,1]$
So, now to get the range of ${\log _e}(\sin x)$
Let us substitute $x = 0$ in the given function ${\log _e}(\sin x)$
i.e.
For $x \to 0 \Rightarrow {\log _e}x = - \infty $
Now let us substitute $x = 1$ in the given function ${\log _e}(\sin x)$
i.e.
For$x \to 1 \Rightarrow {\log _e}x = 0$
Hence from this we can say that the range of given function ${\log _e}(\sin x)$=$( - \infty ,0]$
NOTE: In this particular problem we know the range of $\sin x$ is $[ - 1,1]$ and domain of log function is $(0,1]$ so by substituting the domain values ($x$ values) in the given function i.e. ${\log _e}(\sin x)$ . We will get the range of the given function.
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