Chapter 1, Problem 23RE

To determine

To find: The value of $f^{-1}\left(2\right)$ if $f\left(x\right)=2x+\ln x$.

Answer to Problem 23RE

Solution:

The value of $f^{-1}\left(2\right)=1$.

Explanation of Solution

Given:

The function is, $f\left(x\right)=2x+\ln x$.

Calculation:

Substitute $f\left(x\right)=2$ in the given function and solve as follows.

$\begin{array}{l}2=2x+\ln x\\ 2=\ln e^{2x}+\ln x\left[\because \ln e^x=x\right]\\ 2=\ln \left(e^{2x}\cdot x\right)\\ e^2=e^{\ln \left(x{e}^{2x}\right)}\end{array}$

Simplify further as, $e^2=x{e}^{2x}$.

Notice that this is true only when x = 1.

Therefore, $f\left(1\right)=2$.

That is, the required value is, $f^{-1}\left(2\right)=1$.