What is the answer to e^{2-3x}=5-3x ?

The answer to e^{2-3x}=5-3x is x=(\W_{-1}(-1/(e^3))+5)/3 ,x=(\W_{0}(-1/(e^3))+5)/3

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e^{2-3x}=5-3x

e^{2 - 3 x} = 5 - 3 x

x = \frac{W _{- 1} ( - \frac{1}{e ^{3}} ) + 5}{3}, x = \frac{W _{0} ( - \frac{1}{e ^{3}} ) + 5}{3}

Decimal Notation

x = 0.16491 \dots, x = 1.64917 \dots

Solution steps

e^{2 - 3 x} = 5 - 3 x

Prepare

e^{2 - 3 x} = 5 - 3 x

for Lambert form:

1 = (5 - 3 x) e^{- 2 + 3 x}

Rewrite the equation with

3 x - 5 = u

and

x = \frac{u + 5}{3}

1 = (5 - 3 (\frac{u + 5}{3})) e^{- 2 + 3 (\frac{u + 5}{3})}

Simplify

1 = - e^{u + 3} u

Rewrite

1 = - e^{u + 3} u

in Lambert form:

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

Solve

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

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

Substitute back

u = 3 x - 5,

solve for

x

Solve

3 x - 5 = W_{- 1} (- \frac{1}{e ^{3}}) : x = \frac{W _{- 1} ( - \frac{1}{e ^{3}} ) + 5}{3}

Solve

3 x - 5 = W_{0} (- \frac{1}{e ^{3}}) : x = \frac{W _{0} ( - \frac{1}{e ^{3}} ) + 5}{3}

x = \frac{W _{- 1} ( - \frac{1}{e ^{3}} ) + 5}{3}, x = \frac{W _{0} ( - \frac{1}{e ^{3}} ) + 5}{3}

2^{x} + 2^{x + 1} = 6

100 = 24 e^{5 k}

45000000 = 25700000 \cdot 1. 7^{t}

1 6^{- n} = (\frac{1}{4})^{3 n + 1}

\frac{1}{5 ^{- 3 x}} = 2 5^{x - 1}

What is the answer to e^{2-3x}=5-3x ?
The answer to e^{2-3x}=5-3x is x=(\W_{-1}(-1/(e^3))+5)/3 ,x=(\W_{0}(-1/(e^3))+5)/3