Question 1150779
Question 1150780
<pre>{{{matrix(8,1,
16=6^x,
2^4=(2*3)^x,
2^4=2^x*3^x,
ln(2^4)=ln(2^x*3^x),
4ln(2)=ln(2^x)+ln(3^x),
4ln(2)=xln(2)+x*ln(3),
4ln(2)=x(ln(2)+ln(3)^""),
(4ln(2))/(ln(2)+ln(3))=x )}}}     {{{matrix(9,1,
27=12^y,
3^3=(4*3)^y,
3^3=4^y*3^y,
3^3=(2^2)^y*3^y,
3^3=2^(2y)*3^y,
ln(3^3)=ln(2^(2y))+ln(3^y),
3ln(3)=2y*ln(2)+y*ln(3),
3ln(3)=y(2*ln(2)+ln(3)^""),
(3ln(3))/(2*ln(2)+ln(3))=y )}}}

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

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

Substitute in

{{{matrix(4,1,
(x+a)(y+b),

(((4P)/(P+Q))+a)(((3Q)/(2P+Q))+b),

((4P+aP+aQ)/(P+Q))((3Q+bQ+2bP)/(2P+Q)^""),

(((4+a)P+aQ)/(P+Q)^"")(((3+b)Q+2bP)/(2P+Q)^"") )}}}


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

{{{matrix(7,1,
(((4+a)P+aQ)/(P+Q))(((3+b)Q+2bP)/(2P+Q)),

(((4+4)P+4Q)/(P+Q))(((3+3)Q+2(3)P)/(2P+Q)),

((8P+4Q)/(P+Q)^"")((6Q+6P)/(2P+Q)^""),

((4(2P+Q))/(P+Q)^"")((6(Q+P))/(2P+Q)^""),

((4(cross(2P+Q)))/(cross(P+Q)))((6(cross(Q+P)))/(cross(2P+Q))),

4*6,
24

 


 )}}}

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

Edwin</pre>