How do you simplify #1/2 (log_bM + log_bN - log_bP)#?

1 Answer
Jan 22, 2018

#= log_b sqrt((MN)/P#

Explanation:

Since our bases are all equal we can use #log_b A + log_b B = log_b(AB)# and #log_b A - log_b B = log_b (A/B)#.

#= 1/2log_b((MN)/P)#

We can now use #alogn = logn^a#.

#= log_b((MN)/P)^(1/2)#

#= log_b sqrt((MN)/P)#

Hopefully this helps!