
How do I find the inverse of ${e^x}$?
Answer
540.3k+ views
Hint:Let us try to solve this question in which we have to find the inverse of function ${e^x}$. Before solving this question we will first recall the definition of one-one and onto function, since we have to prove ${e^x}$ to one-one and onto function then only it can have inverse only.
Complete step by step solution:
One-one function: A function $f:X \to Y$ is defined to be one-one if $\forall \,{x_1},{x_2}\, \in \,X,\,f({x_1}) = \,f({x_2})\,\, \Rightarrow \,\,{x_1} = {x_2}$
Onto function: A function $f :X \to Y$ is defined to be onto if$\forall \,y \in Y\,\,,\,\,\exists \,\,x \in X\,such\,that\,f(x) = y$.
Now, we have essential tools to find the inverse of ${e^x}$. Let’s find inverse,
To prove: $f :\Re \to (0,\infty )$defined by$f(x) = {e^x}$ is a one-one function.
Proof: Suppose for every${x_1},{x_2}$, we have$f({x_1}) = f({x_2})$. We will prove that then ${x_1} = {x_2}$.
$
f({x_1}) = f({x_2}) \\
\,\,\,\,\,\,{e^{{x_1}}} = {e^{{x_2}}} \\
$
As we know that from laws of exponents, if ${a^b} = {a^c}$ then $b = c$, using this property, we have
${x_1} = {x_2}$
Hence proved, ${e^x}$ is a one-one function.
To prove: $f:\Re \to (0,\infty )$defined by$f(x) = {e^x}$ is onto function.
Proof: To prove onto function we will find for every $y$ there exists a$x$. Let$y = f(x)$,
$y = f(x) = {e^x}$ $eq(1)$
Taking, natural logarithmic function on both sides of$eq(1)$, we get
$\ln y = x$
Function$f:\Re \to (0,\infty )$ defined by $f(x) = {e^x}$ is onto because $f(y) = \ln \,y\,$for all $y > 0$.
Hence, ${e^x}$ is onto function.
Hence the function $f :\Re \to (0,\infty )$defined by$f(x) = {e^x}$ is one-one and onto function means the inverse exists and the inverse of the function $f(x) = {e^x}$ is $\ ln (x)$.
Note: ${e^x}$ function has an inverse if it is defined from $\Re \to (0,\infty )$. If it is defined from $\Re \to \Re $ then its inverse does not exist because the function will not be onto and the logarithmic function is not defined at $x = 0$. Whenever we are asked to find the inverse of a function we will first try to prove that a given function is one-one and onto.
Complete step by step solution:
One-one function: A function $f:X \to Y$ is defined to be one-one if $\forall \,{x_1},{x_2}\, \in \,X,\,f({x_1}) = \,f({x_2})\,\, \Rightarrow \,\,{x_1} = {x_2}$
Onto function: A function $f :X \to Y$ is defined to be onto if$\forall \,y \in Y\,\,,\,\,\exists \,\,x \in X\,such\,that\,f(x) = y$.
Now, we have essential tools to find the inverse of ${e^x}$. Let’s find inverse,
To prove: $f :\Re \to (0,\infty )$defined by$f(x) = {e^x}$ is a one-one function.
Proof: Suppose for every${x_1},{x_2}$, we have$f({x_1}) = f({x_2})$. We will prove that then ${x_1} = {x_2}$.
$
f({x_1}) = f({x_2}) \\
\,\,\,\,\,\,{e^{{x_1}}} = {e^{{x_2}}} \\
$
As we know that from laws of exponents, if ${a^b} = {a^c}$ then $b = c$, using this property, we have
${x_1} = {x_2}$
Hence proved, ${e^x}$ is a one-one function.
To prove: $f:\Re \to (0,\infty )$defined by$f(x) = {e^x}$ is onto function.
Proof: To prove onto function we will find for every $y$ there exists a$x$. Let$y = f(x)$,
$y = f(x) = {e^x}$ $eq(1)$
Taking, natural logarithmic function on both sides of$eq(1)$, we get
$\ln y = x$
Function$f:\Re \to (0,\infty )$ defined by $f(x) = {e^x}$ is onto because $f(y) = \ln \,y\,$for all $y > 0$.
Hence, ${e^x}$ is onto function.
Hence the function $f :\Re \to (0,\infty )$defined by$f(x) = {e^x}$ is one-one and onto function means the inverse exists and the inverse of the function $f(x) = {e^x}$ is $\ ln (x)$.
Note: ${e^x}$ function has an inverse if it is defined from $\Re \to (0,\infty )$. If it is defined from $\Re \to \Re $ then its inverse does not exist because the function will not be onto and the logarithmic function is not defined at $x = 0$. Whenever we are asked to find the inverse of a function we will first try to prove that a given function is one-one and onto.
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