How do you solve $log(3x)=log(5)+log(x-4)$ ?

Question

5310 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-01T19:13:23

$x=10$

Using the laws of logs we can simplify the right hand side to :

$log(3x)=log[(5)(x-4)]$

$therefore log(3x)=log(5x-20)$

Now take the antilog base 10 on both sides to obtain

$3x=5x-20$

$therefore x=20/2=10$

How do you solve log(3x)=log(5)+log(x-4)log(3x)=log(5)+log(x-4)?