What is the answer to f^5=5^f ?

The answer to f^5=5^f is f=5,f=-(5\W_{0}(-(ln(5))/5))/(ln(5))

The solution is f=5,f=-(5\W_{0}(-(ln(5))/5))/(ln(5))

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f^5=5^f

f^{5} = 5^{f}

f = 5, f = - \frac{5 W _{0} ( - \frac{l n ( 5 )}{5} )}{ln ( 5 )}

Solution steps

f^{5} = 5^{f}

Prepare

f^{5} = 5^{f}

for Lambert form:

f e^{- \frac{l n ( 5 ) f}{5}} = 1

Rewrite the equation with

- \frac{ln ( 5 ) f}{5} = u

and

f = - \frac{5 u}{ln ( 5 )}

(- \frac{5 u}{ln ( 5 )}) e^{u} = 1

Rewrite

(- \frac{5 u}{ln ( 5 )}) e^{u} = 1

in Lambert form:

e^{u} u = - \frac{ln ( 5 )}{5}

Solve

e^{u} u = - \frac{ln ( 5 )}{5} : u = - ln (5), u = W_{0} (- \frac{ln ( 5 )}{5})

u = - ln (5), u = W_{0} (- \frac{ln ( 5 )}{5})

Substitute back

u = - \frac{ln ( 5 ) f}{5},

solve for

f

Solve

- \frac{ln ( 5 ) f}{5} = - ln (5) : f = 5

Solve

- \frac{ln ( 5 ) f}{5} = W_{0} (- \frac{ln ( 5 )}{5}) : f = - \frac{5 W _{0} ( - \frac{l n ( 5 )}{5} )}{ln ( 5 )}

f = 5, f = - \frac{5 W _{0} ( - \frac{l n ( 5 )}{5} )}{ln ( 5 )}

2 (3.1)^{x} = (6.6)^{x}

m^{x + 1} = \frac{1}{2}

2 5^{4 x + 1} = 5^{x - 5}

97 \cdot 0.9 5^{t} = 35

2 \cdot e^{- 6 x} = 95

What is the answer to f^5=5^f ?
The answer to f^5=5^f is f=5,f=-(5\W_{0}(-(ln(5))/5))/(ln(5))
Solve for f: f^5=5^f
The solution is f=5,f=-(5\W_{0}(-(ln(5))/5))/(ln(5))