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

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

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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x^3=2^x

x^{3} = 2^{x}

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

Decimal Notation

x = 9.93953 \dots, x = 1.37346 \dots

Solution steps

x^{3} = 2^{x}

Prepare

x^{3} = 2^{x}

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Rewrite

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

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 )}

3^{x + 5} + 3^{x} = 177876

a^{x} = x

3^{m + 2} = 4

2^{2 x - 2} - 3 \cdot 2^{x} + 8 = 0

e^{(x^{2} - 2 x)} = 1

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