x-Z-n-R-n-x-2n-Does-x-exist-

Question Number 4037 by Filup last updated on 27/Dec/15

\exists x \in Z \exists n \in R : n^{x} = 2 n

Does x e x i s t ?

Commented by Filup last updated on 27/Dec/15

n = x = 2

\Rightarrow 2^{2} = 2 \times 2

Any others ? Can we prove / disprove ?

Answered by RasheedSindhi last updated on 27/Dec/15

\log (n^{x}) = \log (2 n)

x \log n = \log 2 + \log n

x = \frac{\log 2 + \log n}{\log n}

x = 1 + \frac{\log 2}{\log n}

x = 1 + \log_{n} 2

f o r n = 2, x = 2

let n = \sqrt{2}

x = 1 + \log_{\sqrt{2}} 2 = 1 + 2 = 3

{(\sqrt{2})}^{3} = 2 (\sqrt{2})

f o r n =^{m} \sqrt{2}, x = m + 1, m \in N

{(^{m} \sqrt{2})}^{m + 1} = 2 (^{m} \sqrt{2})

Verification :

{(^{m} \sqrt{2})}^{m} .^{m} \sqrt{2} = 2 (^{m} \sqrt{2})

2 (^{m} \sqrt{2}) = 2 (^{m} \sqrt{2})

Commented by Filup last updated on 27/Dec/15

What does this mean ?

Commented by RasheedSindhi last updated on 27/Dec/15

For all n = 2^{\frac{1}{m}}, w h e r e m i s a n y

n a t u r a l, T h e r e e x i s t x = m + 1

a n i n t e g e r s u c h t h a t

{(2^{1 / m})}^{m + 1} = 2 (2^{1 / m})