Question

How do you find the inverse of \[f\left( x \right)={{e}^{-x}}\]?

Answer 1

Hint: This question is from the topic of pre-calculus. In this question, we will find the inverse of the term given in the question. In solving this question, we will first remove the term \[f\left( x \right)\] and replace that term with y in the given equation. After that, we will interchange the terms x and y. After that, we will find the value of y in terms of x. That value will be our answer that is the inverse of \[{{e}^{-x}}\].

Complete step by step answer:
Let us solve this question.
In this question, we have asked to find the inverse of \[f\left( x \right)={{e}^{-x}}\].
Let us write the equation \[f\left( x \right)={{e}^{-x}}\] as
\[y={{e}^{-x}}\]
Now, for finding the inverse, we will first replace the term x with y and replace the term y with x.
So, we can write the equation \[y={{e}^{-x}}\] as
\[x={{e}^{-y}}\]
Now, we will find the value of y in terms of x.
By taking \[\ln \] (or, we can say log base e that is \[{{\log }_{e}}\]) to the both side of above equation, we can write the above equation as
\[\ln x=\ln {{e}^{-y}}\]
Now, using the formula of logarithms: \[\ln {{a}^{b}}=b\ln a\], we can write the above equation as
\[\Rightarrow \ln x=-y\ln e\]
Now, using the formula of logarithms: \[\ln e=1\], we can write the above equation as
\[\Rightarrow \ln x=-y\]
The above equation can also be written as
\[\Rightarrow -y=\ln x\]
Now, multiplying the negative to the both side of equation, we can write the above equation as
\[\Rightarrow -\left( -y \right)=-\ln x\]
As we know that negative multiplied by negative is always positive, so we can write the above equation as
\[\Rightarrow y=-\ln x\]
Using the formula: \[\ln {{a}^{b}}=b\ln a\], we can write the above equation as
\[\Rightarrow y=\ln {{\left( x \right)}^{-1}}\]
Now, using the formula: \[{{\left( x \right)}^{-1}}=\dfrac{1}{x}\], we can write the above equation as
\[\Rightarrow y=\ln \dfrac{1}{x}\]
So, from here we can say that the inverse of \[f\left( x \right)={{e}^{-x}}\] is \[{{f}^{-1}}\left( x \right)=\ln \dfrac{1}{x}\].

Note: We should have a better knowledge in the topic of pre-calculus to solve this type of question easily. We should remember the following formulas:
\[{{\left( x \right)}^{-1}}=\dfrac{1}{x}\]
\[\ln {{a}^{b}}=b\ln a\]
\[\ln e=1\]
Remember the above formulas because they can be helpful in this type of question.