SOLUTION: if 16=6^x, 27=(12)^y find three intger numbers (a, b, c,) satisfy. (x+a) (y+b) =c

     

Let P = ln(2) and Q = ln(3)

So we have (4P)/(P+Q)=x )}}} 

Substitute in




To make that come out an integer, all the P's and Q's must cancel out,
for they are irrational natural logs -- plus, both denominators must 
cancel into both numerators.

The coefficient of the P in the right denominator is twice
the coefficient of Q.

Therefore, the right denominator will cancel into the left numerator if
the coefficient 4+a is twice the coefficient a. So to have that,
4+a = 2a
  4 = a

The coefficient of both P and Q in the left denominator are equal,

Therefore, the left denominator will cancel into the right numerator if 
coefficients 3+b and 2b are equal.  So to have that,
3+b = 2b
  3 = b

So substituting a = 4 and b = 3



So a = 3, b = 4, and c = 24

Edwin