91Ó°ÊÓ

$\text{Thegivenequationis}\ln x+\ln \left(x+2\right)=4$.

We have to express irrational solutions in exact form and as a decimal rounded to three decimal places.

Also, we have to verify the result using the graphing utility.

$\begin{array}{l}\text{Now,}\\ ⇒\ln x+\ln \left(x+2\right)=4\\ ⇒\ln x+\ln \left(x+2\right)=\ln e^4\left[âˆ\mu a=\ln e^a\right]\\ ⇒\ln \left(xâ‹\dots \left(x+2\right)\right)=\ln e^4\left[âˆ\mu \ln a+\ln b=\ln \left(aâ‹\dots b\right)\right]\\ ⇒xâ‹\dots \left(x+2\right)=e^4\\ ⇒x^2+2x=e^4\\ ⇒x^2+2x−e^4=0\end{array}$

$\begin{array}{l}\text{Wewillsolvethisquadraticequationusingformula:}\\ x_{1,2}=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}\\ \text{where}a,b\text{and}c\text{arecoefficientsofequation}a{x}^2+bx+c=0.\\ ⇒x_{1,2}=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}\\ ⇒x_{1,2}=\frac{−2Â\pm \sqrt{4−4â‹\dots 1â‹\dots \left(−e^4\right)}}{2â‹\dots 1}\\ ⇒x_{1,2}=\frac{−2Â\pm \sqrt{4\left(1+e^4\right)}}{2}\\ ⇒x_{1,2}=\frac{−2Â\pm 2\sqrt{1+e^4}}{2}\\ ⇒x_{1,2}=−1Â\pm \sqrt{1+e^4}\\ ⇒x_2=−1+\sqrt{1+e^4}≈6.456\end{array}$

The logarithmic function is well defined only if the argument of the logarithmic function is a positive real number $\left({\log }_a\left(x\right),x>0\right).$

$$\begin{gathered} x>0 \\ x+2>0→x>−2 \\ x∈\left(\right(0,+\mathrm{∞})âˆ\copyright (−2,+\mathrm{∞}\left)\right) \\ ⇒x∈(0,+\mathrm{∞}) \end{gathered}$$

Hence,

The solution of the equation is$x=\left\{−1+\sqrt{1+e^4}\right\}=\left\{6.456\right\}$as$\left\{−1+\sqrt{1+e^4}\right\}=\left\{6.456\right\}∈(0,+\mathrm{∞})$

Plot the graph of the equation using the graphing utility as shown below.

From the above graph. we can observe that the graph of the given equation intersects the x-axis at $x=\left\{−1+\sqrt{1+e^4}\right\}=\left\{6.456\right\}$.