How do you solve #log_8 5 -log_8(x-6)=1#?

1 Answer
Sep 21, 2016

Answer:

#x=53/8 or 6 5/8#

Explanation:

Remembering that
#color(white)("XXX")color(blue)(log_b(p) - log_b(q)=log_b (p/q))#
and that
#color(white)("XXX"color(magenta)(log_b(b) = 1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)(log_8 5 - log_8 (x-6))=color(magenta)(1)#

can be rewritten as
#color(white)("XXX")color(blue)(log_8(5/(x-6))) =color(magenta)( log_8 8)#

which implies
#color(white)("XXX")5/(x-6)=8#

#color(white)("XXX")rarr5=8x-48#

#color(white)("XXX")rarr 8x=53#

#color(white)("XXX")rarr x=53/8#