We have to solve: e^(3x-4) - 1/e^(x-12)=0

e^(3x-4) - 1/e^(x-12)=0

=> e^(3x-4) - e^(-(x-12)) = 0

=> e^(3x-4) = e^(-(x-12))

take the logarithm to the base e of both the sides

=> ln(e^(3x-4)) = ln(e^(-(x-12)))

use the property that ln a^x = x*ln a and ln e = 1

=> 3x - 4 = -(x - 12)

=> 3x - 4 = -x +12

=> 4x = 16

=> x = 4

**The solution of the equation is x = 4**