Question 925847
<pre>
{{{log((5x)) = log((3))+log((x-2))}}}

The base is 10 anytime the base is not written.
But actually it doesn't matter what the base is.
We just need to know some rules of logaritms

Rule 1:   {{{log((A))+log((B))=log((A*B))}}}

Rule 2:  If {{{log((A))=log((B))}}} then we can drop the logs and write {{{A=B}}}

Start with 

{{{log((5x)) = log((3))+log((x-2))}}}

We use rule 1 on the right side:

{{{log((5x)) = log((3(x-2)))}}}

Now use rule 2

{{{5x=3(x-2)}}}

That's an easy equation to solve.  You should get -3

Trouble is, you must always check your answer to see if it
is an actual solution or just an extraneous answer.

{{{log((5x)) = log((3))+log((x-2))}}}

{{{log((5(-3))) = log((3))+log((-3-2))}}}
{{{log((-15)) = log((3))+log((-5))}}}

Oh oh!  -3 is an extraneous answer because there is no such
number in real number mathematics as the logarithm of a negative
number.  Since when we checked we got a logarithm of a negative
number, we knew that the answer we got was extraneous and not a
real solution.

THERE IS NO SOLUTION!

Edwin</pre>