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.
