Chapter 11.6, Problem 57E

To determine

Solve the equation.

Answer to Problem 57E

  $\left(0,-1.0986\right)$ .

Explanation of Solution

Given information:

  $e^{-2x}-4e^{-x}+3=0$

Calculation:

Consider the given equation,

  $e^{-2x}-4e^{-x}+3=0$

  $\begin{array}{l}{e}^{-2x}-4e^{-x}+3=0\\ e^{-2x}-3e^{-x}-e^{-x}+3=0\\ e^{-x}\left(e^{-x}-3\right)-1\left(e^{-x}-3\right)=0\left[factorization\right]\\ \left(e^{-x}-3\right)\left(e^{-x}-1\right)=0\\ \left(e^{-x}-3\right)=0,\left(e^{-x}-1\right)=0\\ e^{-x}=3,e^{-x}=1\\ -x=\ln 3or-x=\ln 1\\ -x=1.0986or-x=0\\ x=-1.0986,x=0\\ x=\left(0,-1.0986\right)\end{array}$

Hence,the value of is $\left(0,-1.0986\right)$ .