What is the answer to solvefor x,e^{zx}=xyz ?

The answer to solvefor x,e^{zx}=xyz is x=-(\W(-1/y))/z

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solvefor x,e^{zx}=xyz

solve for

x, e^{z x} = x yz

x = - \frac{W ( - \frac{1}{y} )}{z}

Solution steps

e^{z x} = x yz

Prepare

e^{z x} = x yz

for Lambert form:

1 = x yz e^{- z x}

Rewrite the equation with

- z x = u

and

x = - \frac{u}{z}

1 = (- \frac{u}{z}) yz e^{u}

Rewrite

1 = (- \frac{u}{z}) yz e^{u}

in Lambert form:

e^{u} u = - \frac{1}{y}

Solve

e^{u} u = - \frac{1}{y} : u = W (- \frac{1}{y})

u = W (- \frac{1}{y})

Substitute back

u = - z x,

solve for

x

Solve

- z x = W (- \frac{1}{y}) : x = - \frac{W ( - \frac{1}{y} )}{z}

x = - \frac{W ( - \frac{1}{y} )}{z}

\frac{71}{3} = 3^{x}

x, 15 - 0. 5^{x} = - 6

5^{x} = - 5

5^{x} = - 7

3^{2 x - 5} = 1

What is the answer to solvefor x,e^{zx}=xyz ?
The answer to solvefor x,e^{zx}=xyz is x=-(\W(-1/y))/z