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**

We'll begin by applying the negative power property of exponentials:

1/e^(x-12) = e^-(x-12)

Now, we'll re-write the equation:

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

Since the bases are matching, we'll use one to one property and we'll equate the exponents:

3x - 4 = -x + 12

We'll isolate x to the left side.

3x + x - 4 = 12

We'll combine like terms and we'll add 4 both sides:

4x = 12 + 4

4x = 16

We'll divide by 4:

x = 4

The root of the exponential equation is x = 4.

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