we have the equation
[tex]ln(x)+ln(x+8)=4[/tex]
Applying property of log
[tex]lnx*(x+8)=4[/tex][tex]\begin{gathered} e^4=x(x+8) \\ e^4=x^2+8x \\ x^2+8x-e^4=0 \end{gathered}[/tex]
Find out the exact solution
Solve the quadratic equation using the formula
a=1
b=8
c=-e^4
substitute
[tex]x=\frac{-8\pm\sqrt{8^2-4(1)(-e^4)}}{2(1)}[/tex][tex]x=\frac{-8\pm\sqrt{64+4e^4}}{2}[/tex]
Simplify
[tex]x=\frac{-8\pm2\sqrt{16+e^4}}{2}[/tex]
therefore the exact solutions are
[tex]\begin{gathered} x=-4+\sqrt{16+e^4} \\ x=-4-\sqrt{16+e^4} \end{gathered}[/tex]
The decimal approximation of the solutions are
[tex]\begin{gathered} x=4.402 \\ x=-12.402 \end{gathered}[/tex]
