What is the answer to (nx)^x=n^{n^n} ?

The answer to (nx)^x=n^{n^n} is x=(n^nln(n))/(\W(n^{n+1)ln(n))}

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(nx)^x=n^{n^n}

(n x)^{x} = n^{n^{n}}

x = \frac{n ^{n} ln ( n )}{W ( n ^{n + 1} ln ( n ) )}

Solution steps

(n x)^{x} = n^{n^{n}}

Prepare

(n x)^{x} = n^{n^{n}}

for Lambert form:

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

Rewrite the equation with

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

and

x = \frac{n ^{n} ln ( n )}{u + ln ( n )}

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

Solve

e^{- u} (\frac{n ^{n} ln ( n )}{u + ln ( n )}) = 1 : u = W (n^{n + 1} ln (n)) - ln (n)

u = W (n^{n + 1} ln (n)) - ln (n)

Substitute back

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

solve for

x

Solve

- (1 - \frac{n ^{n}}{x}) ln (n) = W (n^{n + 1} ln (n)) - ln (n) : x = \frac{n ^{n} ln ( n )}{W ( n ^{n + 1} ln ( n ) )}

x = \frac{n ^{n} ln ( n )}{W ( n ^{n + 1} ln ( n ) )}

e^{2 x} = (e^{x})^{2}

e^{x} = 14

e^{x} = 25

e^{x} = 3 x

(\frac{1}{2})^{1 - x} = 4

What is the answer to (nx)^x=n^{n^n} ?
The answer to (nx)^x=n^{n^n} is x=(n^nln(n))/(\W(n^{n+1)ln(n))}