How do you condense #(1/2)log_6 9 +log_6 5#?

Question

1620 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-02T01:19:14

I tried this:

We can write:
#log_6(9)^(1/2)+log_6(5)=#
#=log_6(sqrt(9))+log_6(5)=#
#=log_6(3)+log_6(5)=#
#=log_6(3*5)=log_6(15)#

Answer 2 · 2016-06-02T01:25:09

#log_6 15=log 15 /log 6=1.5114#, nearly.

Use #n log_b a = log_b (a^n), log_b(mn)=log_b m+log_b n,and log _b a = log_c a/log_c b#.

So, #(1/2)log_6 9 + log_6 5#

#=log_6(9^(1/2)) + log_6 5#

#=log_6 (3X5)=log_6 15#

#=log 15/log 6=1.5114#, nearly