Given that the 2x power of 3 = the 3x power of 4 = the 6th power of 12, the basic problem of finding the value of 3 / x + 2 / y is higher than one

Given that the 2x power of 3 = the 3x power of 4 = the 6th power of 12, the basic problem of finding the value of 3 / x + 2 / y is higher than one

Let 3 ^ 2x = 4 ^ 3Y = 12 ^ 6 = K
Then log3 (k) = 2x, log4 (k) = 3Y, log12 (k) = 6 (1)
Deform (1)
logk(3)=1/2X,logk(4)=1/3Y,logk(12)=1/6
logk(3)+logk(4)=1/2x+1/3y
That is logK (12) = 1 / 2x + 1 / 3Y = 1 / 6, the denominator is (3Y + 2x) / 6xy
And because: 3 / x + 2 / y = (3Y + 2x) / XY
The difference between the two formulas is one sixth
That is, (3Y + 2x) / 6xy * 6 is equal to the required formula
That is, 1 / 6 * 6 = 1.0