If #log_b x = 2/3log_b 27 + 2 log_b 2 -log_b 3#, what is #x#?

1 Answer
Sep 19, 2016

Answer:

#x=color(green)(12)#

Explanation:

Things you need to know and remember:
#color(white)("XXX")log_b (p^q) = q * log_b p#
and
#color(white)("XXX")log_b (p) + log_b (q) = log_b (pq)#

#color(red)(2/3log_b 27)#
#color(white)("XXX")=2/3 log_b 3^3 = 2/3 * 3 log_b 3 = color(red)(2 log_b 3)#

#color(blue)(2log_b 2)#
#color(white)("XXX")=log_b 2^2 = color(blue)(log_b 4)#

#color(red)(2/3 log_b 27) + color(blue)(2 log_b 4) - color(black)(log_b 3)#
#color(white)("XXX")=color(red)(2log_b 3) +color(blue)(log_b 4) -color(black)(log_b 3)#

#color(white)("XXX")=log_b 3 + log_b 4#

#color(white)("XXX")=log_b 12#

Therefore
#color(white)("XXX")log_b x = 2/3log_b 27 + 2 log_b 2 -log_b 3#

#color(white)("XXX")rArr log_b x = log_b 12#

#color(white)("XXX")rArr x =12#