Note that
#color(white)("XXX")log_color(blue)(8) color(red)(8) =1# since #color(blue)(8)^1=color(red)(8)#
Combining this with the given equation we have:
#color(white)("XXX")log_8(x-5)+(log_8(x+2)=log_8 8#
Using the log product rule:
#color(white)("XXX")log_8 ((x-5)(x+2)) = log_8 8#
#rArr#
#color(white)("XXX")(x-5)(x+2)=8#
#color(white)("XXX")x^2-3x-10=8#
#color(white)("XXX")x^2-3x-18 = 0#
#color(white)("XXX")(x-6)(x+3)=0#
#rArr x=6# or #x=-3#
Since the #log# function is not defined for negative arguments
#x=-3# must be an extraneous solution.
