What is the answer to solvefor a,x(a+b)=x^{ab} ?

The answer to solvefor a,x(a+b)=x^{ab} is a=-(\W(-\frac{bln(x))/(x^{b^2+1)})+b^2ln(x)}{bln(x)}

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solvefor a,x(a+b)=x^{ab}

solve for

a, x (a + b) = x^{ab}

a = - \frac{W ( - \frac{b l n ( x )}{x ^{b^{2} + 1}} ) + b ^{2} ln ( x )}{b ln ( x )}

Solution steps

x (a + b) = x^{ab}

Prepare

x (a + b) = x^{ab}

for Lambert form:

e^{l n (x) (1 - ab)} (a + b) = 1

Rewrite the equation with

(- ab - b^{2}) ln (x) = u

and

a = - \frac{u + b ^{2} ln ( x )}{b ln ( x )}

e^{l n (x) (1 - (- \frac{u + b ^{2} l n ( x )}{b l n ( x )}) b)} (- \frac{u + b ^{2} ln ( x )}{b ln ( x )} + b) = 1

Simplify

e^{l n (x) + u + b^{2} l n (x)} (- \frac{u + b ^{2} ln ( x )}{b ln ( x )} + b) = 1

Rewrite

e^{l n (x) + u + b^{2} l n (x)} (- \frac{u + b ^{2} ln ( x )}{b ln ( x )} + b) = 1

in Lambert form:

e^{u} u = - \frac{b ln ( x )}{x ^{b^{2} + 1}}

Solve

e^{u} u = - \frac{b ln ( x )}{x ^{b^{2} + 1}} : u = W (- \frac{b ln ( x )}{x ^{b^{2} + 1}})

u = W (- \frac{b ln ( x )}{x ^{b^{2} + 1}})

Substitute back

u = (- ab - b^{2}) ln (x),

solve for

a

Solve

(- ab - b^{2}) ln (x) = W (- \frac{b ln ( x )}{x ^{b^{2} + 1}}) : a = - \frac{W ( - \frac{b l n ( x )}{x ^{b^{2} + 1}} ) + b ^{2} ln ( x )}{b ln ( x )}

a = - \frac{W ( - \frac{b l n ( x )}{x ^{b^{2} + 1}} ) + b ^{2} ln ( x )}{b ln ( x )}

9^{x^{2} - x} = 81

3 e^{- 2 x} = 5

(- 2)^{x} = 15

10 + e^{- x} = 0

2 5^{2 x + 4} = 125 \cdot 5^{x - 1}

What is the answer to solvefor a,x(a+b)=x^{ab} ?
The answer to solvefor a,x(a+b)=x^{ab} is a=-(\W(-\frac{bln(x))/(x^{b^2+1)})+b^2ln(x)}{bln(x)}