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):

$48=4p$

$p=12$

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