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

1 Answer
Gió
Jul 30, 2015

I found: #b=+-4#

Explanation:

You can start using a property of logs:
#logx+logy=logxy#
so you get:
#log_6[2(b^2+2)]=2#
using the definition of log:
#log_ax=b -> x=a^b# you get:
#2(b^2+2)=6^2=36#
#b^2+2=36/2=18#
#b^2=18-2=16#
so that #b=+-sqrt(16)=+-4#

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