What is the answer to e^{x-4}+2.5x-11.7=0 ?

The answer to e^{x-4}+2.5x-11.7=0 is x=-(2.5\W_{0}(\frac{e^{0.68})/(2.5))-11.7}{2.5}

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e^{x-4}+2.5x-11.7=0

e^{x - 4} + 2.5 x - 11.7 = 0

x = - \frac{2.5 W _{0} ( \frac{e ^{0.68}}{2.5} ) - 11.7}{2.5}

Decimal Notation

x = 4.19424 \dots

Solution steps

e^{x - 4} + 2.5 x - 11.7 = 0

Prepare

e^{x - 4} + 2.5 x - 11.7 = 0

for Lambert form:

1 = - e^{- x + 4} (2.5 x - 11.7)

Rewrite the equation with

- x + \frac{11.7}{2.5} = u

and

x = - u + 4.68

1 = - e^{- (- u + 4.68) + 4} (2.5 (- u + 4.68) - 11.7)

Simplify

1 = 2.5 e^{u - 0.68} u

Rewrite

1 = 2.5 e^{u - 0.68} u

in Lambert form:

e^{u} u = \frac{e ^{0.68}}{2.5}

Solve

e^{u} u = \frac{e ^{0.68}}{2.5} : u = W_{0} (\frac{e ^{0.68}}{2.5})

u = W_{0} (\frac{e ^{0.68}}{2.5})

Substitute back

u = - x + \frac{11.7}{2.5},

solve for

x

Solve

- x + \frac{11.7}{2.5} = W_{0} (\frac{e ^{0.68}}{2.5}) : x = - \frac{2.5 W _{0} ( \frac{e ^{0.68}}{2.5} ) - 11.7}{2.5}

x = - \frac{2.5 W _{0} ( \frac{e ^{0.68}}{2.5} ) - 11.7}{2.5}

k, n^{l o g_{10} (n)} = n^{k}

x, 2^{x} = 6

2^{(1 - x^{2})} = \frac{1}{8}

2^{x + 1} = 4

1 6^{x} = 0.5

What is the answer to e^{x-4}+2.5x-11.7=0 ?
The answer to e^{x-4}+2.5x-11.7=0 is x=-(2.5\W_{0}(\frac{e^{0.68})/(2.5))-11.7}{2.5}