log_6(3x)+log_6(x-1)=3 Solve the equation. Check for extraneous solutions.

To evaluate the given equation log_6(3x)+log_6(x-1)=3 , we may apply the logarithm property: log_b(x)+log_b(y)=log_b(x*y) .
log_6(3x)+log_6(x-1)=3
log_6(3x*(x-1))=3
log_6(3x^2-3x)=3
To get rid of the "log" function, we may apply the logarithm property: b^(log_b(x))=x .
Raise both sides by base of 6 .
6^(log_6(3x^2-3x))=6^3
3x^2-3x=216
Subtract 216 from both sides of the equation to simplify in standard form: ax^2+bx+c= 0 .
3x^2-3x-216=216-216
3x^2-3x-216=0
Apply factoring on the trinomial.
3*(x + 8)*(x - 9)=0
Apply zero-factor property to solve for x by equating each factor in terms of x to 0 .
x+8-8=0-8
x=-8
 
x-9=0
x-9+9=0+9
x=9
Checking: Plug-in each x on  log_6(3x)+log_6(x-1)=3 .
Let x=-8 on log_6(3x)+log_6(x-1)=3 .
log_6(3*(-8))+log_6(-8-1)=?3
log_6(-24)+log_6(-9)=?3
undefined +undefined =?3     FALSE
Note that log_b(x) is undefined on xlt=0 .
Let x=9 on log_6(3x)+log_6(x-1)=3 .
log_6(3*9)+log_6(9-1)=?3
log_6(27)+log_6(8)=?3
log_6(27*8)=?3
log_6(216)=?3
log_6(6^3)=?3
3log_6(6)=?3
3*1=?3
3=3    TRUE
 
Thus, the x=-8 is an extraneous solution.
The x=9 is the only real solution of the equation log_6(3x)+log_6(x-1)=3 .