Chapter 6, Problem 33CR

To determine

To solve : the equation

Answer to Problem 33CR

The solution of equation is $x=\sqrt{e^3+1}+1$

Explanation of Solution

Given information The equation is $\ln x+\ln x+\ln 2=3$

Calculation :

The solution of equation is

  $\begin{array}{l}\ln x+\ln x+\ln 2=3\\ \ln x+\ln \left(x+2\right)=3\\ \ln x\left(x+2\right)=3\\ x\left(x+2\right)=3\\ x=e^3\\ x^2+2x=e^3\\ x^2+2x+1=e^3+1\\ {\left(x+2\right)}^2=e^3+1\\ \left(x+1\right)=\sqrt[\pm ]{e^3+1}+1\end{array}$

Therefore, by taking negative sign: $x=-\sqrt{e^3+1}-1$

Putting the value of $x$ in equation $\ln x+\ln x+\ln 2=3$

  $\ln \left(\sqrt[\pm ]{e^3+1}-1\right)+\ln \left(\sqrt[\pm ]{e^3+1}-1+2\right)=3$

The logarithm defines for positive values or argument but here argument is negative so the logarithm can’t define.

Therefore. $x=-\sqrt{e^3+1}+1$ is an extraneous

Taking positive sign

  $x=\sqrt{e^3+1}+1$

Putting the value of $x$ in equation $\ln x+\ln x+\ln 2=3$

  $\begin{array}{l}\ln \left(\sqrt{e^3+1}-1\right)+\ln \left(\sqrt{e^3+1}-1+2\right)=3\\ \ln \left(\sqrt{e^3+1}-1\right)+\ln \left(\sqrt{e^3+1}+1\right)=3\\ \ln \left(\sqrt{e^3+1}-1\right)\left(\sqrt{e^3+1}+1\right)=3\\ \left(\sqrt{e^3+1}-1\right)\left(\sqrt{e^3+1}+1\right)=e^3\\ \left(\sqrt{e^3+1}\right)\left(\sqrt{e^3+1}\right)-\sqrt{e^3+1}+\sqrt{e^3+1}-1=e^3\\ 1=1\end{array}$

Hence, the solution of equation is $x=\sqrt{e^3+1}+1$