What is the answer to 2^x-x^3=0 ?

The answer to 2^x-x^3=0 is x=-(3\W_{-1}(-(ln(2))/3))/(ln(2)),x=-(3\W_{0}(-(ln(2))/3))/(ln(2))

Solve for x: 2^x-x^3=0

The solution is x=-(3\W_{-1}(-(ln(2))/3))/(ln(2)),x=-(3\W_{0}(-(ln(2))/3))/(ln(2))

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2^{x} - x^{3} = 0

x = - \frac{3 W _{- 1} ( - \frac{l n ( 2 )}{3} )}{ln ( 2 )}, x = - \frac{3 W _{0} ( - \frac{l n ( 2 )}{3} )}{ln ( 2 )}

Solution steps

2^{x} - x^{3} = 0

Prepare

2^{x} - x^{3} = 0

for Lambert form:

1 = x e^{- \frac{l n ( 2 ) x}{3}}

Rewrite the equation with

- \frac{ln ( 2 ) x}{3} = u

and

x = - \frac{3 u}{ln ( 2 )}

1 = (- \frac{3 u}{ln ( 2 )}) e^{u}

Rewrite

1 = (- \frac{3 u}{ln ( 2 )}) e^{u}

in Lambert form:

e^{u} u = - \frac{ln ( 2 )}{3}

Solve

e^{u} u = - \frac{ln ( 2 )}{3} : u = W_{- 1} (- \frac{ln ( 2 )}{3}), u = W_{0} (- \frac{ln ( 2 )}{3})

u = W_{- 1} (- \frac{ln ( 2 )}{3}), u = W_{0} (- \frac{ln ( 2 )}{3})

Substitute back

u = - \frac{ln ( 2 ) x}{3},

solve for

x

Solve

- \frac{ln ( 2 ) x}{3} = W_{- 1} (- \frac{ln ( 2 )}{3}) : x = - \frac{3 W _{- 1} ( - \frac{l n ( 2 )}{3} )}{ln ( 2 )}

Solve

- \frac{ln ( 2 ) x}{3} = W_{0} (- \frac{ln ( 2 )}{3}) : x = - \frac{3 W _{0} ( - \frac{l n ( 2 )}{3} )}{ln ( 2 )}

x = - \frac{3 W _{- 1} ( - \frac{l n ( 2 )}{3} )}{ln ( 2 )}, x = - \frac{3 W _{0} ( - \frac{l n ( 2 )}{3} )}{ln ( 2 )}

6 (5^{x - 4}) = 42

9^{x} = 2 7^{3 - x}

2 3^{n} = - (5^{n})

7^{x} - 7 = 35 + 7^{3 - x}

What is the answer to 2^x-x^3=0 ?
The answer to 2^x-x^3=0 is x=-(3\W_{-1}(-(ln(2))/3))/(ln(2)),x=-(3\W_{0}(-(ln(2))/3))/(ln(2))
Solve for x: 2^x-x^3=0
The solution is x=-(3\W_{-1}(-(ln(2))/3))/(ln(2)),x=-(3\W_{0}(-(ln(2))/3))/(ln(2))