What is the answer to e^x=2x+3 ?

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

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e^x=2x+3

e^{x} = 2 x + 3

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

Decimal Notation

x = 1.92393 \dots, x = - 1.37337 \dots

Solution steps

e^{x} = 2 x + 3

Prepare

e^{x} = 2 x + 3

for Lambert form:

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

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})}

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

e^{u} u = - \frac{1}{2 e ^{\frac{3}{2}}} : u = W_{- 1} (- \frac{1}{2 e ^{\frac{3}{2}}}), u = W_{0} (- \frac{1}{2 e ^{\frac{3}{2}}})

u = W_{- 1} (- \frac{1}{2 e ^{\frac{3}{2}}}), u = W_{0} (- \frac{1}{2 e ^{\frac{3}{2}}})

Substitute back

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

solve for

x

Solve

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

Solve

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

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

2^{3 x - 4} = 10

(5 y)^{y} = 5^{5^{5}}

5^{2 x} = 625

e^{3 x} = 15

e^{3 x} = 19

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