What is the answer to 2^n=n^2 ?

The answer to 2^n=n^2 is n=-(2\W_{0}((ln(2))/2))/(ln(2)),n=2,n=4

The solution is n=-(2\W_{0}((ln(2))/2))/(ln(2)),n=2,n=4

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2^n=n^2

2^{n} = n^{2}

n = - \frac{2 W _{0} ( \frac{l n ( 2 )}{2} )}{ln ( 2 )}, n = 2, n = 4

Decimal Notation

n = - 0.76666 \dots, n = 2, n = 4

Solution steps

2^{n} = n^{2}

Take both sides of the equation to the power of

\frac{1}{2}

(2^{n})^{\frac{1}{2}} = (n^{2})^{\frac{1}{2}}

Simplify

(2^{n})^{\frac{1}{2}} : 2^{\frac{n}{2}}

Simplify

(n^{2})^{\frac{1}{2}} : ∣ n ∣

Solve

2^{\frac{n}{2}} = ∣ n ∣ : n = - \frac{2 W _{0} ( \frac{l n ( 2 )}{2} )}{ln ( 2 )} or n = 2 or n = 4

n = - \frac{2 W _{0} ( \frac{l n ( 2 )}{2} )}{ln ( 2 )}, n = 2, n = 4

x, e^{x} y + y = x - 1

4^{x} (5^{2 x}) = 10

850 (1 - e^{- 0.06 t}) = 600

5^{x} - 2 = \frac{62 3 ^{3}}{5 ^{2}}

x^{2 x} + 16 = 8 x^{x}

What is the answer to 2^n=n^2 ?
The answer to 2^n=n^2 is n=-(2\W_{0}((ln(2))/2))/(ln(2)),n=2,n=4
Solve for n: 2^n=n^2
The solution is n=-(2\W_{0}((ln(2))/2))/(ln(2)),n=2,n=4