What is the answer to 9e^{(-x)/2}+12e^{(-x)/6}=0 ?

The answer to 9e^{(-x)/2}+12e^{(-x)/6}=0 is x=Undefined

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9e^{(-x)/2}+12e^{(-x)/6}=0

9 e^{\frac{- x}{2}} + 12 e^{\frac{- x}{6}} = 0

x = Undefined

Solution steps

9 e^{\frac{- x}{2}} + 12 e^{\frac{- x}{6}} = 0

Subtract

12 e^{\frac{- x}{6}}

from both sides

9 e^{\frac{- x}{2}} + 12 e^{\frac{- x}{6}} - 12 e^{\frac{- x}{6}} = 0 - 12 e^{\frac{- x}{6}}

Simplify

9 e^{\frac{- x}{2}} = - 12 e^{\frac{- x}{6}}

Apply exponent rules

ln (9) + \frac{- x}{2} = ln (- 12) + \frac{- x}{6}

Solve

ln (9) + \frac{- x}{2} = ln (- 12) + \frac{- x}{6} : x =

Undefined

x = Undefined

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2^{3000} = 1 0^{x}

(\frac{1}{2})^{n} + 1 = 2

What is the answer to 9e^{(-x)/2}+12e^{(-x)/6}=0 ?
The answer to 9e^{(-x)/2}+12e^{(-x)/6}=0 is x=Undefined