How do you solve #log_6(b^2+2) + log_6 2=2#?

1 Answer
Jun 20, 2015

#b=+-4#

Explanation:

Start by usig the following property of logs:

#log_ab+log_ac=log_a(b*c)#
so you get:

#log_6[2(b^2+2)]=2#

now use the definition of log:
#log_ax=b ->x=a^b#

in your case:
#2(b^2+2)=6^2#
#b^2+2=18#
#b^2=16#
#b=+-sqrt(16)=+-4#