What is the answer to x^7=7^x ?

The answer to x^7=7^x is x=7,x=-(7\W_{0}(-(ln(7))/7))/(ln(7))

The solution is x=7,x=-(7\W_{0}(-(ln(7))/7))/(ln(7))

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x^7=7^x

x^{7} = 7^{x}

x = 7, x = - \frac{7 W _{0} ( - \frac{l n ( 7 )}{7} )}{ln ( 7 )}

Solution steps

x^{7} = 7^{x}

Prepare

x^{7} = 7^{x}

for Lambert form:

x e^{- \frac{l n ( 7 ) x}{7}} = 1

Rewrite the equation with

- \frac{ln ( 7 ) x}{7} = u

and

x = - \frac{7 u}{ln ( 7 )}

(- \frac{7 u}{ln ( 7 )}) e^{u} = 1

Rewrite

(- \frac{7 u}{ln ( 7 )}) e^{u} = 1

in Lambert form:

e^{u} u = - \frac{ln ( 7 )}{7}

Solve

e^{u} u = - \frac{ln ( 7 )}{7} : u = - ln (7), u = W_{0} (- \frac{ln ( 7 )}{7})

u = - ln (7), u = W_{0} (- \frac{ln ( 7 )}{7})

Substitute back

u = - \frac{ln ( 7 ) x}{7},

solve for

x

Solve

- \frac{ln ( 7 ) x}{7} = - ln (7) : x = 7

Solve

- \frac{ln ( 7 ) x}{7} = W_{0} (- \frac{ln ( 7 )}{7}) : x = - \frac{7 W _{0} ( - \frac{l n ( 7 )}{7} )}{ln ( 7 )}

x = 7, x = - \frac{7 W _{0} ( - \frac{l n ( 7 )}{7} )}{ln ( 7 )}

e^{1 - 2 x} = e^{3 x - 5}

8^{2 x + 3} = 2^{a} b^{x}

e^{5 l n (x)} = 32

(5)^{3 x - 1} = 4

2^{3 x} = 7

What is the answer to x^7=7^x ?
The answer to x^7=7^x is x=7,x=-(7\W_{0}(-(ln(7))/7))/(ln(7))
Solve for x: x^7=7^x
The solution is x=7,x=-(7\W_{0}(-(ln(7))/7))/(ln(7))