What is the answer to e^2x-e^x=0 ?

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

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e^2x-e^x=0

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

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

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

- x = u

and

x = - u

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

Simplify

- 1 = e^{2 + u} u

Rewrite

- 1 = e^{2 + u} u

in Lambert form:

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

Solve

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

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

Substitute back

u = - x,

solve for

x

Solve

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

Solve

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

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

3 \cdot b - e^{b} = 0

6^{x} = 45

3^{2 x + 1} = 7

\frac{27}{1 + e ^{- x}} = 2

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