Question 905391
<pre>
log(6x-3)base 3=1+log(x-3)base3)

What the other person who responded, says (see below), is TOTALLY FALSE!! His/Her answer is also WRONG, so IGNORE it all!!

WRONG!! WRONG!! WRONG!!
Since the bases of the logs are the same (number 3 in this case), then the insides must be equal. That is:

6x - 3 = 1 + x - 3
6x - x = 3 - 3 + 1
5x = 1
x = 1/5
Understand?

{{{log(3, (6x - 3)) = 1 + log (3, (x - 3))}}}
Notice that there is a 1 ATTACHED to the log on the right! 
Anyway, the smaller variable-expression, x - 3, MUST be > 0. We then get: x - 3 > 0, and x > 3. The problem then becomes:
{{{log(3, (6x - 3)) = 1 + log (3, (x - 3))}}}, with the constraint, x > 3.
{{{log(3, (6x - 3)) - log (3, (x - 3)) = 1}}}
         {{{log(3, ((6x - 3)/(x - 3))) = 1}}} ---- Applying {{{log (b, (a)) - log (b, (c))}}} = {{{log (b, (a/c))}}}
                {{{(6x - 3)/(x - 3) = 3^1}}} --- Converting to EXPONENTIAL form             
                6x - 3 = 3(x - 3) ----- Cross-multiplying
                6x - 3 = 3x - 9 
               6x - 3x = - 9 + 3
                    3x = - 6
                     {{{x = (- 6)/3 = - 2}}}
However, x = - 2 is NOT a solution, as it is NOT > 3, which makes this x-value, EXTRANEOUS!. So, there are <font color = red><font size = 4><b>NO SOLUTIONS!</font></font></b></pre>