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