How do you solve for x in #log_6x=2-log_6 4#?

Question

2397 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-03T14:09:44

#color(blue)(x=9)#

#log_6(x)=2-log_6(4)#

#log_6(x)+log_6(4)=2#

#log(a)+log(b)=log(ab)#

#log_6(4x)=2#

Raising the base to these:

#6^(log_6(4x))=6^2#

#4x=6^2#

#x=(6^2)/4=36/4=9#