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Question: Find the inverse of the function,\(y = {3^x}\) and check if the inverse is also a function....

Find the inverse of the function,y=3xy = {3^x} and check if the inverse is also a function.

Explanation

Solution

Inverse of a given one to one function f(x)f(x) is defined as, g(x)g(x). Here g(x)g(x) is equal to f−1(x){f^{ - 1}}(x). We may also say that if f(x)f(x) and g(x)g(x) are two one-to-one functions, then (fog)(x)=x(fog)(x) = x. Also, to be kept in mind that, f−1(x)≠1f(x){f^{ - 1}}(x) \ne \dfrac{1}{{f(x)}}.

Complete step by step solution:
Let us assume, f(x)=yf(x) = y.
And, we know from the question that y=3xy = {3^x}
Replace every x with y and every y with x.
Thus, f(y)=x=3yf(y) = x = {3^y}……. equation(i)
Solving for y, we have

log⁡(f(y))=log⁡3y ⇒log⁡(x)=ylog⁡3 ⇒y=log⁡(x)log⁡3 \log (f(y)) = \log {3^y} \\\ \Rightarrow \log (x) = y\log 3 \\\ \Rightarrow y = \dfrac{{\log (x)}}{{\log 3}} \\\

The resultant function in y be named f−1(x){f^{ - 1}}(x).
This function f−1(x){f^{ - 1}}(x) is the inverse of the function, f(x)f(x). To verify the same, we need to put the values in these equations and tally the answers.
(f∘f−1)(x)=x\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = x…….. equation(iii)
(f−1∘f)(x)=x\left( {{f^{ - 1}} \circ f} \right)\left( x \right) = x…….. equation(iv)
Equation (iii) can be written as,
(f∘f−1)(x)=f[f−1(x)]\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = f[{f^{ - 1}}(x)]
Putting the values of f−1(x){f^{ - 1}}(x) in the above equation,
⇒(f∘f−1)(x)=f[log⁡(x)log⁡3]\Rightarrow \left( {f \circ {f^{ - 1}}} \right)\left( x \right) = f[\dfrac{{\log (x)}}{{\log 3}}]

x$$……….(Hence the equation is verified.) Next, we need to verify equation (iv), $$\left( {{f^{ - 1}} \circ f} \right)\left( x \right) = {f^{ - 1}}[f(x)]$$ Putting the values of in the above equation, $$ \Rightarrow \left( {{f^{ - 1}} \circ f} \right)\left( x \right) = {f^{ - 1}}[{3^x}]$$ $$ \Rightarrow \left( {{f^{ - 1}} \circ f} \right)\left( x \right) = \dfrac{{\log [{3^x}]}}{{\log 3}} = x$$……….. (Hence the equation is verified.) **As the definition of an inverse function says, two one-to-one functions are inverse of each other if they satisfy the equations $$\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = x$$and $$\left( {{f^{ - 1}} \circ f} \right)\left( x \right) = x$$, and hence the found inverse of the given function is also a function.** **Note:** Let $f$ be a function whose domain is called set X, and whose codomain is called set Y. Then $f$ is invertible if there exists a function g with domain Y and image (range) X, with the property: $f(x) = y{\mkern 1mu} {\mkern 1mu} \Leftrightarrow {\mkern 1mu} {\mkern 1mu} g(y) = x.$ If $f$ is invertible, then the function $$g$$ is exclusive, which implies that there 's exactly one function $$g$$ satisfying this property. That function $$g$$ is then called the inverse of $f$, and is typically denoted as${f^{ - 1}}$.