How do you solve the equation #log_5 64-log_5 (8/3)+log_5 2=log_5 (4p)#?

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Answer 1 · 2016-11-15T13:50:43

#p=12#

We have:

#log_5(64)-log_5(8/3)+log_5(2)=log_5(4p)#

Let's first combine the left side:

#log_5((64xx2)/(8/3))=log_5(4p)#

#log_5((64xx2xx3)/(8))=log_5(4p)#

#log_5(8xx2xx3)=log_5(4p)#

#log_5(48)=log_5(4p)#

We can now take the inverse function (effectively getting rid of the logs):

#48=4p#

#p=12#