SOLUTION: log(6x-3)base 3=1+log(x-3)base3)

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Question 905391: log(6x-3)base 3=1+log(x-3)base3)
Found 2 solutions by harpazo, MathTherapy:
Answer by harpazo(655) About Me  (Show Source):
You can put this solution on YOUR website!
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?


Answer by MathTherapy(10699) About Me  (Show Source):
You can put this solution on YOUR website!
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%283%2C+%286x+-+3%29%29+=+1+%2B+log+%283%2C+%28x+-+3%29%29
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%283%2C+%286x+-+3%29%29+=+1+%2B+log+%283%2C+%28x+-+3%29%29, with the constraint, x > 3.
log%283%2C+%286x+-+3%29%29+-+log+%283%2C+%28x+-+3%29%29+=+1
         log%283%2C+%28%286x+-+3%29%2F%28x+-+3%29%29%29+=+1 ---- Applying log+%28b%2C+%28a%29%29+-+log+%28b%2C+%28c%29%29 = log+%28b%2C+%28a%2Fc%29%29
                %286x+-+3%29%2F%28x+-+3%29+=+3%5E1 --- Converting to EXPONENTIAL form             
                6x - 3 = 3(x - 3) ----- Cross-multiplying
                6x - 3 = 3x - 9 
               6x - 3x = - 9 + 3
                    3x = - 6
                     x+=+%28-+6%29%2F3+=+-+2
However, x = - 2 is NOT a solution, as it is NOT > 3, which makes this x-value, EXTRANEOUS!. So, there are NO SOLUTIONS!