Chapter 5 - 5.4 - Exponential and Logarithmic Equations - 5.4 Exercises - Page 395: 60

$x=2$

$\log_2x+\log_2(x+2)=\log_2(x+6)$ $\log_2[x(x+2)]=\log_2(x+6)$ $\log_2(x^2+2x)=\log_2(x+6)$ Using the One-to-One Property: $x^2+2x=x+6$ $x^2+x-6=0~~$ (Make: $x=3x-2x$) $x^2+3x-2x-6=0$ $x(x+3)-2(x+3)=0$ $(x-2)(x+3)=0$ $x-2=0$ $x=2$ $x+3=0$ $x=-3$. But, the $-3$ does not lie in the domain of $\ln x$. So, it is not a valid solution.