Option c: x=6 is a true solution and x=-6 is an extraneous solution.
Explanation:
The equation is 
Dividing both sides of the equation by 2, we get,

Simplifying,

Since, we know by the logarithmic definition, if
, then 
Using this definition, we have,

Hence, 
Now, let us verify if
is the solution.
Substitute
in the equation
to see whether both sides of the equation are true.
We have,

Using the log rule, 
We have,

Hence, both sides of the equation are equal.
Thus,
is the true solution.