What is the answer to 3e^{3x}-7e^{2x}-19e^x-5+4e^{-x}=0 ?

The answer to 3e^{3x}-7e^{2x}-19e^x-5+4e^{-x}=0 is x=-ln(3),x=2ln(2)

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3e^{3x}-7e^{2x}-19e^x-5+4e^{-x}=0

3 e^{3 x} - 7 e^{2 x} - 19 e^{x} - 5 + 4 e^{- x} = 0

x = - ln (3), x = 2 ln (2)

Decimal Notation

x = - 1.09861 \dots, x = 1.38629 \dots

Solution steps

3 e^{3 x} - 7 e^{2 x} - 19 e^{x} - 5 + 4 e^{- x} = 0

Apply exponent rules

3 (e^{x})^{3} - 7 (e^{x})^{2} - 19 e^{x} - 5 + 4 (e^{x})^{- 1} = 0

Rewrite the equation with

e^{x} = u

3 (u)^{3} - 7 (u)^{2} - 19 u - 5 + 4 (u)^{- 1} = 0

Solve

3 u^{3} - 7 u^{2} - 19 u - 5 + 4 u^{- 1} = 0 : u = - 1, u = \frac{1}{3}, u = 4

u = - 1, u = \frac{1}{3}, u = 4

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = - 1 :

No Solution for

x \in R

Solve

e^{x} = \frac{1}{3} : x = - ln (3)

Solve

e^{x} = 4 : x = 2 ln (2)

x = - ln (3), x = 2 ln (2)

3^{x} - 3^{- x} = \frac{728}{27}

\frac{17}{327} = e^{- 9.22 \times 1 0^{- 4 t}}

7^{3 x + 4} = 4 9^{2 x + 1}

y, (3^{3})^{- 1} = x^{y}

e^{x + 9} = \frac{1}{e}

What is the answer to 3e^{3x}-7e^{2x}-19e^x-5+4e^{-x}=0 ?
The answer to 3e^{3x}-7e^{2x}-19e^x-5+4e^{-x}=0 is x=-ln(3),x=2ln(2)