Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 25E from Chapter 1.5 from Stewart's Calculus, 8th Edition.
Problem 25E
Chapter:
Problem:
Find a formula for the inverse of the function.
Step-by-Step Solution
Given information
We are given with following function\[y = \ln \left( {x + 3} \right)\]We have to find the inverse of this function. TO find the inverse of a function y, we have to write the dependent variable $x$ in terms of y and replace x with inverse function and $y$ with $x$
Step 1: Apply Exponential and simplify
\[\begin{array}{l}y = \ln \left( {x + 3} \right)\\{e^y} = {e^{\ln \left( {x + 3} \right)}}\\{e^y} = x + 3\,\,\,\,\,\left\{ {\,{\rm{Since}}\,\,\,{e^{\ln \left( {ax} \right)}} = ax} \right\}\\x + 3\, = {e^y}\\x = {e^y} - 3\end{array}\]
Step 2: Replace x and y
\[\begin{array}{l}x = {e^y} - 3\\y = {e^x} - 3\end{array}\]Therefore, the inverse is \[y = {e^x} - 3\]
We are given with following function\[y = \ln \left( {x + 3} \right)\]We have to find the inverse of this function. TO find the inverse of a function y, we have to write the dependent variable $x$ in terms of y and replace x with inverse function and $y$ with $x$
Step 1: Apply Exponential and simplify
\[\begin{array}{l}y = \ln \left( {x + 3} \right)\\{e^y} = {e^{\ln \left( {x + 3} \right)}}\\{e^y} = x + 3\,\,\,\,\,\left\{ {\,{\rm{Since}}\,\,\,{e^{\ln \left( {ax} \right)}} = ax} \right\}\\x + 3\, = {e^y}\\x = {e^y} - 3\end{array}\]
Step 2: Replace x and y
\[\begin{array}{l}x = {e^y} - 3\\y = {e^x} - 3\end{array}\]Therefore, the inverse is \[y = {e^x} - 3\]