Question

How do you find the inverse of $y = {e^{x - 1}}$ ?

Answer 1

Here we will use the following instruction to find the inverse of an exponential function with domain $( - \infty ,\infty )$ and range $(0,\infty )$.we first write the function as an equation as follows $y = {e^{x - 1}}$ ‘given exponential function’ , change the variable $x$ into $y$ and $y$ into $x$ and take the natural logarithm of both sides to obtain the inverse function.
Formula used :
$y = {f^{ - 1}}(x) \Leftrightarrow f(y)$

Complete step by step answer: Given function $y = {e^{x - 1}}$ this is a one to one function,
Let us first find the domain and range of the given function,
Domain of $y:( - \infty ,\infty )$ and Range: for $x$ in the domain, the range of ${e^{x - 1}}$ is given by$(0,\infty )$
To compute the inverse we need to follow the following steps,
Step-1:
Switch the variable $x$ for $y$ and $y$ for $x$ like this,
$x = {e^{y - 1}}$
Step-2:
Begin to solve for $y$, take $\log$ on both side, remember that $\log x$ is the inverse function for ${e^x}$ which means that both $\log \left( {{e^x}} \right) = x$ and ${e^{\log x}} = x$ hold. This means that we can apply $\log$on both sides of the equation to “get rid” of the exponential function.
$\Rightarrow \log \left( x \right) = \log \left( {{e^{y - 1}}} \right)$
Step-3:
Using the properties of $\log$, we know that ,
$\log e = 1$ and $\log {a^n} = n\log a$ we get,
$\Rightarrow \log x = y - 1$
$\Rightarrow 1 + \log x = y$
Step-4:
Now just replace $y$ with ${f^{ - 1}}(x)$ to obtain the inverse function we get,
$\Rightarrow {f^{ - 1}}(x) = 1 + \log x$
Hence the inverse of the function$y = {e^{x - 1}}$
$\Rightarrow {f^{ - 1}}(x) = 1 + \log x$

Note:
The domain and range of the inverse function are respectively the range and domain of the given function.
Hence the domain and range of ${f^{ - 1}}$ are given by the domain $(0,\infty )$ and range$( - \infty ,\infty )$.
To solve exponential function we use logarithm vice versa.
Remember: one to one function is the only function that has an inverse that is a function.