What is the answer to (n-2)^{(n-2)2}=2 ?

The answer to (n-2)^{(n-2)2}=2 is n=(ln(2)}{\W_{0}(\frac{ln(2))/2)*2}+2

Solve for n: (n-2)^{(n-2)2}=2

The solution is n=(ln(2)}{\W_{0}(\frac{ln(2))/2)*2}+2

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

n = \frac{ln ( 2 )}{W _{0} ( \frac{l n ( 2 )}{2} ) \cdot 2} + 2

Solution steps

(n - 2)^{(n - 2) \cdot 2} = 2

Prepare

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

for Lambert form:

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

Rewrite the equation with

\frac{ln ( 2 )}{2 ( n - 2 )} = u

and

n = \frac{ln ( 2 )}{u \cdot 2} + 2

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

Solve

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

u = W_{0} (\frac{ln ( 2 )}{2})

Substitute back

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

solve for

n

Solve

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

n = \frac{ln ( 2 )}{W _{0} ( \frac{l n ( 2 )}{2} ) \cdot 2} + 2

2 x^{4} e^{- 2 x} = 0

4^{x + 3} = 64

1.7 \cdot 6^{- 1.7 v} - 6 = 8

3^{x} = \frac{1}{243}

0. 1^{2 x} = 10

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