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

The answer to solvefor x,e^f(x)+f-x=1 is x=(1-f)/(e^f-1);\quad f\ne 0

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solvefor x,e^f(x)+f-x=1

solve for

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

x = \frac{1 - f}{e ^{f} - 1}; f \neq = 0

Solution steps

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

e^{f} x + f - x = 1

Move

f

to the right side

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

Factor

e^{f} x - x : x (e^{f} - 1)

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

Divide both sides by

e^{f} - 1; f \neq = 0

x = \frac{1 - f}{e ^{f} - 1}; f \neq = 0

x, \frac{x}{2} + \frac{( 3 y )}{5} = - 1

a, 160 a + 100 b = 64

a, 72.48 = a (24 + \frac{b}{100})

x, d^{2} + 2 d + 3 y = sinh (x)

x = \frac{- 7}{11}

What is the answer to solvefor x,e^f(x)+f-x=1 ?
The answer to solvefor x,e^f(x)+f-x=1 is x=(1-f)/(e^f-1);\quad f\ne 0