SOLUTION: Log6x+log6(x-5)=2 Check for inadmissible roots

Log6x+log6(x-5)=2
Check for inadmissible roots
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Respondent, nycsub_teacher(90) provides the following solution,
"Log6x+log6(x-5)=2
We write it like this:
log_6(x/(x-5) = 2
x/(x-5) = 6^2
x/(x-5) = 36," and then asks, "Can you finish now?"

Whether or not the person can finish doesn't really matter, because the solution that'll be derived from the respondent's attempt, will be WRONG!!

Correct way: 
The smaller of the 2 log-variable arguments, x - 5 MUST be > 0, and so, x - 5 > 0____x > 5
So, we get: , with x > 5
                                  ---- Applying  = 
                                             -- Converting from LOGARITHMIC form to EXPONENTIAL form 
                                           
                                    
                                 (x - 9)(x + 4) = 0 ---- FACTORIZING the trinomial
                                  x - 9 = 0      or     x + 4 = 0 ---- Setting each factor equal to 0 (zero)
                                        x = 9      or            x = - 4 (IGNORE)

As one of the 2 values, x = - 4, is NOT > 5, it is an INADMISSIBLE root, or an EXTRANEOUS solution.

Therefore, the SOLE ADMISSIBLE/VALID/ACCEPTABLE solution to this equation is x = 9.