What is the answer to e^{x-3}+1=4x-3 ?

The answer to e^{x-3}+1=4x-3 is x=-\W_{-1}(-1/(4e^2))+1,x=-\W_{0}(-1/(4e^2))+1

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e^{x-3}+1=4x-3

e^{x - 3} + 1 = 4 x - 3

x = - W_{- 1} (- \frac{1}{4 e ^{2}}) + 1, x = - W_{0} (- \frac{1}{4 e ^{2}}) + 1

Solution steps

e^{x - 3} + 1 = 4 x - 3

Prepare

e^{x - 3} + 1 = 4 x - 3

for Lambert form:

1 = (4 x - 4) e^{- x + 3}

Rewrite the equation with

- x + 1 = u

and

x = - u + 1

1 = (4 (- u + 1) - 4) e^{- (- u + 1) + 3}

Simplify

1 = - 4 e^{u + 2} u

Rewrite

1 = - 4 e^{u + 2} u

in Lambert form:

e^{u} u = - \frac{1}{4 e ^{2}}

Solve

e^{u} u = - \frac{1}{4 e ^{2}} : u = W_{- 1} (- \frac{1}{4 e ^{2}}), u = W_{0} (- \frac{1}{4 e ^{2}})

u = W_{- 1} (- \frac{1}{4 e ^{2}}), u = W_{0} (- \frac{1}{4 e ^{2}})

Substitute back

u = - x + 1,

solve for

x

Solve

- x + 1 = W_{- 1} (- \frac{1}{4 e ^{2}}) : x = - W_{- 1} (- \frac{1}{4 e ^{2}}) + 1

Solve

- x + 1 = W_{0} (- \frac{1}{4 e ^{2}}) : x = - W_{0} (- \frac{1}{4 e ^{2}}) + 1

x = - W_{- 1} (- \frac{1}{4 e ^{2}}) + 1, x = - W_{0} (- \frac{1}{4 e ^{2}}) + 1

3 e^{t} = 5 + 8 e^{- t}

8 = 3^{x}

8^{4 x + 2} = 64

x, 1 = e^{x y}

4^{x + 2} = 12 8^{1 - x}

What is the answer to e^{x-3}+1=4x-3 ?
The answer to e^{x-3}+1=4x-3 is x=-\W_{-1}(-1/(4e^2))+1,x=-\W_{0}(-1/(4e^2))+1