How do you express #3log_5 4 - 5log_5 2# as a single logarithm?

1 Answer
Apr 8, 2017

Answer:

#3 log_5 4 - 5 log_5 2 =1/log_2 5#

Explanation:

To solve this, you will need to use the "Change of Base" formula:
#log_a x = (log_b x) / (log_b a)#
where we change the base-#a# logarithm to base-#b#.

Let's see how we can apply this to what we have.
I notice that we have a 4 and a 2 inside a base-5 logarithm.
So maybe I should change to a base-2 logarithm.
Let's see how each log transforms:
#log_5 4 = (log_2 4)/(log_2 5)#
but since #4=2^2#,
#log_2 4 = log_2 2^2= 2#
so
#log_5 4 = 2/(log_2 5)#
and similarly,
#log_5 2 = 1/(log_2 5)#

So, our expression becomes:
#3 log_5 4 - 5 log_5 2 = (3*2)/(log_2 5) - (5*1)/(log_2 5)#
i.e.
#=(6-5)/(log_2 5)#
#=1/log_2 5#

There could be other answers using different bases.