
How do you find the inverse of \[y = {6^x} + 1\] ?
Answer
546.3k+ views
Hint: Here in this question, we have to find the inverse of the given function y or \[f(x)\] . The inverse of a function is denoted by \[{f^{ - 1}}(x)\] . Here first we have to write the function in terms of x and then we have to solve for y using mathematics operations and simplification we get the required solution.
Complete step by step solution:
An inverse function or an anti-function is defined as a function, which can reverse into another function. In simple words, if any function “ \[f\] ” takes \[x\] to \[y\] then, the inverse of “ \[f\] ” will take \[y\] to \[x\] . If the function is denoted by ‘ \[f\] ’ or ‘ \[F\] ’, then the inverse function is denoted by \[{f^{ - 1}}\] or \[{F^{ - 1}}\] .
i.e., If \[f\] and \[g\] are inverse functions, then \[f\left( x \right) = y\] if and only if \[g\left( y \right) = x\] .
Consider the given function
\[y = {6^x} + 1\] --------(1)
switch the \[x\] 's and the \[y\] 's means replace \[x\] as \[y\] and \[y\] as \[x\] . i.e., \[f(x)\] is a substitute for " \[y\] ".
Equation (1) can be written as function of \[x\] i.e.,
\[ \Rightarrow x = {6^y} + 1\] ------(2)
Now, to find the inverse we have to solve the equation (2) for \[y\] .
Subtract 1 on both side
\[ \Rightarrow x - 1 = {6^y} + 1 - 1\]
On simplification, we get
\[ \Rightarrow x - 1 = {6^y}\]
On rearranging
\[ \Rightarrow {6^y} = x - 1\]
Take the logarithm with base 6 on both sides
\[ \Rightarrow {\log _6}\left( {{6^y}} \right) = {\log _6}\left( {x - 1} \right)\]
By the logarithm property \[\ln \left( {{m^n}} \right) = n\ln \left( m \right)\] , then
\[ \Rightarrow y{\log _6}\left( 6 \right) = {\log _6}\left( {x - 1} \right)\]
By the one more property of logarithm with base e is \[{\ln _e}\left( e \right) = 1\] , here we have base 6 the \[{\log _6}\left( 6 \right) = 1\]
\[ \Rightarrow y = {\log _6}\left( {x - 1} \right)\]
Hence, the inverse of a function \[y = {6^x} + 1\] is \[y = {\log _6}\left( {x - 1} \right)\] .
So, the correct answer is “ \[y = {\log _6}\left( {x - 1} \right)\] ”.
Note: In this question we must know about the logarithmic functions as we know that the logarithmic function and exponential function are inverse of each other. While using the logarithmic functions we must know the properties of logarithmic function. We must know about the simple arithmetic operations.
Complete step by step solution:
An inverse function or an anti-function is defined as a function, which can reverse into another function. In simple words, if any function “ \[f\] ” takes \[x\] to \[y\] then, the inverse of “ \[f\] ” will take \[y\] to \[x\] . If the function is denoted by ‘ \[f\] ’ or ‘ \[F\] ’, then the inverse function is denoted by \[{f^{ - 1}}\] or \[{F^{ - 1}}\] .
i.e., If \[f\] and \[g\] are inverse functions, then \[f\left( x \right) = y\] if and only if \[g\left( y \right) = x\] .
Consider the given function
\[y = {6^x} + 1\] --------(1)
switch the \[x\] 's and the \[y\] 's means replace \[x\] as \[y\] and \[y\] as \[x\] . i.e., \[f(x)\] is a substitute for " \[y\] ".
Equation (1) can be written as function of \[x\] i.e.,
\[ \Rightarrow x = {6^y} + 1\] ------(2)
Now, to find the inverse we have to solve the equation (2) for \[y\] .
Subtract 1 on both side
\[ \Rightarrow x - 1 = {6^y} + 1 - 1\]
On simplification, we get
\[ \Rightarrow x - 1 = {6^y}\]
On rearranging
\[ \Rightarrow {6^y} = x - 1\]
Take the logarithm with base 6 on both sides
\[ \Rightarrow {\log _6}\left( {{6^y}} \right) = {\log _6}\left( {x - 1} \right)\]
By the logarithm property \[\ln \left( {{m^n}} \right) = n\ln \left( m \right)\] , then
\[ \Rightarrow y{\log _6}\left( 6 \right) = {\log _6}\left( {x - 1} \right)\]
By the one more property of logarithm with base e is \[{\ln _e}\left( e \right) = 1\] , here we have base 6 the \[{\log _6}\left( 6 \right) = 1\]
\[ \Rightarrow y = {\log _6}\left( {x - 1} \right)\]
Hence, the inverse of a function \[y = {6^x} + 1\] is \[y = {\log _6}\left( {x - 1} \right)\] .
So, the correct answer is “ \[y = {\log _6}\left( {x - 1} \right)\] ”.
Note: In this question we must know about the logarithmic functions as we know that the logarithmic function and exponential function are inverse of each other. While using the logarithmic functions we must know the properties of logarithmic function. We must know about the simple arithmetic operations.
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