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

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

Explanation

Solution

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(x)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{{e}^{-x}}.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to find the inverse of f(x)=e−xf\left( x \right)={{e}^{-x}}.
Let us write the equation f(x)=e−xf\left( x \right)={{e}^{-x}} as
y=e−xy={{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−xy={{e}^{-x}} as
x=e−yx={{e}^{-y}}
Now, we will find the value of y in terms of x.
By taking ln⁡\ln (or, we can say log base e that is log⁡e{{\log }_{e}}) to the both side of above equation, we can write the above equation as
ln⁡x=ln⁡e−y\ln x=\ln {{e}^{-y}}
Now, using the formula of logarithms: ln⁡ab=bln⁡a\ln {{a}^{b}}=b\ln a, we can write the above equation as
⇒ln⁡x=−yln⁡e\Rightarrow \ln x=-y\ln e
Now, using the formula of logarithms: ln⁡e=1\ln e=1, we can write the above equation as
⇒ln⁡x=−y\Rightarrow \ln x=-y
The above equation can also be written as
⇒−y=ln⁡x\Rightarrow -y=\ln x
Now, multiplying the negative to the both side of equation, we can write the above equation as
⇒−(−y)=−ln⁡x\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
⇒y=−ln⁡x\Rightarrow y=-\ln x
Using the formula: ln⁡ab=bln⁡a\ln {{a}^{b}}=b\ln a, we can write the above equation as
⇒y=ln⁡(x)−1\Rightarrow y=\ln {{\left( x \right)}^{-1}}
Now, using the formula: (x)−1=1x{{\left( x \right)}^{-1}}=\dfrac{1}{x}, we can write the above equation as
⇒y=ln⁡1x\Rightarrow y=\ln \dfrac{1}{x}
So, from here we can say that the inverse of f(x)=e−xf\left( x \right)={{e}^{-x}} is f−1(x)=ln⁡1x{{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:
(x)−1=1x{{\left( x \right)}^{-1}}=\dfrac{1}{x}
ln⁡ab=bln⁡a\ln {{a}^{b}}=b\ln a
ln⁡e=1\ln e=1
Remember the above formulas because they can be helpful in this type of question.