Chapter 5 - 5.2 - Logarithmic Functions and Their Graphs - 5.2 Exercises - Page 378: 32

$x=3$ or $x=-9$

$\log(x^2+6x)=\log27$ Using the One-to-One Property: $x^2+6x=27$ $x^2+6x-27=0~~$ ($a=1,~b=6,~c=-27$) $x=\frac{-6±\sqrt {6^2-4(1)(-27)}}{2(1)}=\frac{-6±\sqrt {144}}{2}=\frac{-6±12}{2}=-3±6$ $x=3$ or $x=-9$