What is the answer to solvefor n,x^n+n=0 ?

The answer to solvefor n,x^n+n=0 is n=-(\W(ln(x)))/(ln(x))

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solvefor n,x^n+n=0

solve for

n, x^{n} + n = 0

n = - \frac{W ( ln ( x ) )}{ln ( x )}

Solution steps

x^{n} + n = 0

Prepare

x^{n} + n = 0

for Lambert form:

1 = - n e^{- n l n (x)}

Rewrite the equation with

- n ln (x) = u

and

n = - \frac{u}{ln ( x )}

1 = - (- \frac{u}{ln ( x )}) e^{u}

Rewrite

1 = - (- \frac{u}{ln ( x )}) e^{u}

in Lambert form:

e^{u} u = ln (x)

Solve

e^{u} u = ln (x) : u = W (ln (x))

u = W (ln (x))

Substitute back

u = - n ln (x),

solve for

n

Solve

- n ln (x) = W (ln (x)) : n = - \frac{W ( ln ( x ) )}{ln ( x )}

n = - \frac{W ( ln ( x ) )}{ln ( x )}

x, 4^{x} = 8^{x - 1}

6 = 6^{x}

95 = 2 9^{x}

y, z = x (1.8)^{x y^{2} + y}

5.2 (1.12)^{n} = 8

What is the answer to solvefor n,x^n+n=0 ?
The answer to solvefor n,x^n+n=0 is n=-(\W(ln(x)))/(ln(x))