How do you solve # log_4 a + log_4 5 = 3#?

1 Answer
Alan P.
Feb 28, 2016

#a=12.8#

Explanation:

Things you need to know:
[1]#color(white)("XXX")log_bp+log_bq = log_b(pq)#
[2]#color(white)("XXX")log_b m = k##color(white)("XXXX")# means #color(white)("XXX")b^k=m#

#log_4 a + log_4 5 = log_4 5a = 3# from [1] and given equality

#4^3=5a# from [2]

#rarr 5a = 64#

#rarr a=64/5 = 12.8#

Related questions
See all questions in Logarithmic Models
Impact of this question
1299 views around the world
You can reuse this answer
Creative Commons License