What is the answer to solvefor x,f^{-1}(x)=3+e^{2-x} ?

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

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solvefor x,f^{-1}(x)=3+e^{2-x}

solve for

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

x = W (\frac{f}{e ^{3 f - 2}}) + 3 f

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

x - 3 f = u

and

x = u + 3 f

f = (u + 3 f - 3 f) e^{- 2 + u + 3 f}

Simplify

f = e^{- 2 + u + 3 f} u

Rewrite

f = e^{- 2 + u + 3 f} u

in Lambert form:

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

Solve

e^{u} u = \frac{f}{e ^{3 f - 2}} : u = W (\frac{f}{e ^{3 f - 2}})

u = W (\frac{f}{e ^{3 f - 2}})

Substitute back

u = x - 3 f,

solve for

x

Solve

x - 3 f = W (\frac{f}{e ^{3 f - 2}}) : x = W (\frac{f}{e ^{3 f - 2}}) + 3 f

x = W (\frac{f}{e ^{3 f - 2}}) + 3 f

3 x (1 0^{x}) + 1 0^{x} = 0

3^{2 z} = 81

2^{3 y - 2} = 16

2^{2 x - 3} = 8^{x + 1}

2^{x} = 5^{x}

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