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

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

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e^x+e^2x=e^3x

e^{x} + e^{2} x = e^{3} x

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

Decimal Notation

x = 3.90309 \dots, x = 0.08581 \dots

Solution steps

e^{x} + e^{2} x = e^{3} x

Prepare

e^{x} + e^{2} x = e^{3} x

for Lambert form:

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

Rewrite the equation with

- x = u

and

x = - u

1 = (e - 1) e^{2 - (- u)} (- u)

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

u = - x,

solve for

x

Solve

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

Solve

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

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

4^{x + 1} - 5^{x + 1} = (0.8) (4^{x})

1.73 = 3^{x}

5^{- 3 x - 1} = 25

x^{l o g_{10} (x)} = 100

3^{x} = \frac{1}{729}

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