What is the answer to solvefor y,zx+e^{zy}=x ?

The answer to solvefor y,zx+e^{zy}=x is y=(ln(x-zx))/z

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solvefor y,zx+e^{zy}=x

solve for

y, z x + e^{zy} = x

y = \frac{ln ( x - z x )}{z}

Solution steps

z x + e^{zy} = x

Move

z x

to the right side

e^{zy} = x - z x

Apply exponent rules

zy = ln (x - z x)

Solve

zy = ln (x - z x) : y = \frac{ln ( x - z x )}{z}

y = \frac{ln ( x - z x )}{z}

x, f = m \cdot 6^{- x}

x, y + 2 y = 2 e^{x}

lo g_{3} (8 1^{x}) = 16

3 e^{5 x} = 10

1 + 6 e^{- t} = e^{t}

What is the answer to solvefor y,zx+e^{zy}=x ?
The answer to solvefor y,zx+e^{zy}=x is y=(ln(x-zx))/z