log2 (c-12) +log2 (2c)=7  --->  

(c-12)*(2c) =    --->

(c-12)*c =   --->

 = ,

(C-16)*(c+4) = 0  --->

c = 16  and/or  c = -4 --->

c = 16 is the unique solution which survives under the logarithm.


Answer.  c = 16.

Use the rule of logarithm that says that the sum of
two logs equals the log of the product of what the
two logs are taken of.  That rule is usually written

    .

Therefore



becomes

 

Use the rule of logarithms that says that a logarithm equals
the exponent to which its base must be raised in order
to get the number that the log is being taken of.  That is 
usually written:

    is equivalent to 

Therefore

  becomes








Divide through by 2



Get 0 on the right by subtracting 64 from both sides:





c-16 = 0;   c+4 = 0
   c = 16     c = -4

We must ignore the answer c=-4 as extraneous
because in the original equation:



if we substitute -4 we will have the logarithm of
a negative number which is not defined in ordinary
mathematics.  However the answer 16 is valid.

Edwin