What is the answer to 2^x=x^5 ?

The answer to 2^x=x^5 is x=-(5\W_{-1}(-(ln(2))/5))/(ln(2)),x=-(5\W_{0}(-(ln(2))/5))/(ln(2))

The solution is x=-(5\W_{-1}(-(ln(2))/5))/(ln(2)),x=-(5\W_{0}(-(ln(2))/5))/(ln(2))

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2^x=x^5

Solution

2^{x} = x^{5}

Solution

x = - \frac{5 W _{- 1} ( - \frac{l n ( 2 )}{5} )}{ln ( 2 )}, x = - \frac{5 W _{0} ( - \frac{l n ( 2 )}{5} )}{ln ( 2 )}

Decimal

x = 22.44000 \dots, x = 1.17727 \dots

Solution steps

2^{x} = x^{5}

Prepare

2^{x} = x^{5}

for Lambert form:

1 = x e^{- \frac{l n ( 2 ) x}{5}}

Rewrite the equation with

- \frac{ln ( 2 ) x}{5} = u

and

x = - \frac{5 u}{ln ( 2 )}

1 = (- \frac{5 u}{ln ( 2 )}) e^{u}

Rewrite

1 = (- \frac{5 u}{ln ( 2 )}) e^{u}

in Lambert form:

e^{u} u = - \frac{ln ( 2 )}{5}

Solve

e^{u} u = - \frac{ln ( 2 )}{5} : u = W_{- 1} (- \frac{ln ( 2 )}{5}), u = W_{0} (- \frac{ln ( 2 )}{5})

u = W_{- 1} (- \frac{ln ( 2 )}{5}), u = W_{0} (- \frac{ln ( 2 )}{5})

Substitute back

u = - \frac{ln ( 2 ) x}{5},

solve for

x

Solve

- \frac{ln ( 2 ) x}{5} = W_{- 1} (- \frac{ln ( 2 )}{5}) : x = - \frac{5 W _{- 1} ( - \frac{l n ( 2 )}{5} )}{ln ( 2 )}

Solve

- \frac{ln ( 2 ) x}{5} = W_{0} (- \frac{ln ( 2 )}{5}) : x = - \frac{5 W _{0} ( - \frac{l n ( 2 )}{5} )}{ln ( 2 )}

x = - \frac{5 W _{- 1} ( - \frac{l n ( 2 )}{5} )}{ln ( 2 )}, x = - \frac{5 W _{0} ( - \frac{l n ( 2 )}{5} )}{ln ( 2 )}

Graph

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Frequently Asked Questions (FAQ)

What is the answer to 2^x=x^5 ?
The answer to 2^x=x^5 is x=-(5\W_{-1}(-(ln(2))/5))/(ln(2)),x=-(5\W_{0}(-(ln(2))/5))/(ln(2))
Solve for x: 2^x=x^5
The solution is x=-(5\W_{-1}(-(ln(2))/5))/(ln(2)),x=-(5\W_{0}(-(ln(2))/5))/(ln(2))

2^x=x^5

Solution

Solution

Graph

Popular Examples

Frequently Asked Questions (FAQ)

What is the answer to 2^x=x^5 ?

Solve for x: 2^x=x^5