What is the answer to e^{x-1}-6=-2x-3 ?

The answer to e^{x-1}-6=-2x-3 is x=-(2\W_{0}((e^{1/2})/2)-3)/(2)

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e^{x-1}-6=-2x-3

e^{x - 1} - 6 = - 2 x - 3

x = - \frac{2 W _{0} ( \frac{e ^{\frac{1}{2}}}{2} ) - 3}{2}

Solution steps

e^{x - 1} - 6 = - 2 x - 3

Prepare

e^{x - 1} - 6 = - 2 x - 3

for Lambert form:

1 = (- 2 x + 3) e^{- x + 1}

Rewrite the equation with

- x + \frac{3}{2} = u

and

x = - \frac{2 u - 3}{2}

1 = (- 2 (- \frac{2 u - 3}{2}) + 3) e^{- (- \frac{2 u - 3}{2}) + 1}

Simplify

1 = 2 e^{\frac{2 u - 3}{2} + 1} u

Rewrite

1 = 2 e^{\frac{2 u - 3}{2} + 1} u

in Lambert form:

e^{u} u = \frac{e ^{\frac{1}{2}}}{2}

Solve

e^{u} u = \frac{e ^{\frac{1}{2}}}{2} : u = W_{0} (\frac{e ^{\frac{1}{2}}}{2})

u = W_{0} (\frac{e ^{\frac{1}{2}}}{2})

Substitute back

u = - x + \frac{3}{2},

solve for

x

Solve

- x + \frac{3}{2} = W_{0} (\frac{e ^{\frac{1}{2}}}{2}) : x = - \frac{2 W _{0} ( \frac{e ^{\frac{1}{2}}}{2} ) - 3}{2}

x = - \frac{2 W _{0} ( \frac{e ^{\frac{1}{2}}}{2} ) - 3}{2}

x - x e^{5 x + 2} = 0

3^{x + 1} = 5

e^{- 4 x} - 5 e^{- 2 x} + 6 = 0

5^{3 x + 2} = 2 5^{x - 1}

3^{5 x} = 7^{1 - 5 x}

What is the answer to e^{x-1}-6=-2x-3 ?
The answer to e^{x-1}-6=-2x-3 is x=-(2\W_{0}((e^{1/2})/2)-3)/(2)