exponential equatione^(3x-4) - 1/e^(x-12)=0
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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.