I'll remember some concepts:
[tex]\log_{a}{b} = \log_{a}{c} \implies b=c\\\\\log_{a}{b}+ \log_{a}{c} = \log_{a}{b\cdot c}[/tex]
Question:
[tex]\log_{6}{(x^2-5x)}=\log_{6}{(2x-9)}+\log_{6}({x+3)}\\\\\\\log_{6}{(x^2-5x)}=\log_{6}{(2x-9)\cdot(x+3)}\\\\\\(x^2-5x) = (2x-9)\cdot(x+3)\\\\\\x^2-5x=2x^2+6x-9x-27\\\\\\0 = x^2+2x-27\\\\\\\Delta = 4+108 = 112\\\\\\\\\sqrt{\Delta} = \sqrt{112} = 4\sqrt{7}\\\\\\\\x = \dfrac{-2\pm 4\sqrt{7}}{2}\\\\\\\\x = -1 \pm 2\sqrt{7} \\\\\\\\x_1 = 2\sqrt{7} -1\\\\\\x_2 = -1-2\sqrt{7}[/tex]
The negative solution cant exist since we are talking about logarithms.
