What is the answer to solvefor a,a^b=b^a ?

The answer to solvefor a,a^b=b^a is a=-(\W(-(ln(b))/b)b)/(ln(b))

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solvefor a,a^b=b^a

solve for

a, a^{b} = b^{a}

a = - \frac{W ( - \frac{l n ( b )}{b} ) b}{ln ( b )}

Solution steps

a^{b} = b^{a}

Prepare

a^{b} = b^{a}

for Lambert form:

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

Rewrite the equation with

- \frac{a ln ( b )}{b} = u

and

a = - \frac{b u}{ln ( b )}

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

Rewrite

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

in Lambert form:

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

Solve

e^{u} u = - \frac{ln ( b )}{b} : u = W (- \frac{ln ( b )}{b})

u = W (- \frac{ln ( b )}{b})

Substitute back

u = - \frac{a ln ( b )}{b},

solve for

a

Solve

- \frac{a ln ( b )}{b} = W (- \frac{ln ( b )}{b}) : a = - \frac{W ( - \frac{l n ( b )}{b} ) b}{ln ( b )}

a = - \frac{W ( - \frac{l n ( b )}{b} ) b}{ln ( b )}

x, 4^{x} + 2^{(x + 2)} = 32

4 \cdot e^{- 7 a - 5} = - 9

23 \cdot 3^{2 x} = 115

(3^{x})^{3} = 2 7^{x}

\frac{( 2 ^{n} )}{n + 1} = 51.2

What is the answer to solvefor a,a^b=b^a ?
The answer to solvefor a,a^b=b^a is a=-(\W(-(ln(b))/b)b)/(ln(b))