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

∃x∈Z∃n∈R:n^x =2n Does x exist?

$$\exists{x}\in\mathbb{Z}\exists{n}\in\mathbb{R}:{n}^{{x}} =\mathrm{2}{n} \\ $$$$\mathrm{Does}\:{x}\:{exist}? \\ $$

Commented by Filup last updated on 27/Dec/15

n=x=2 ⇒2^2 =2×2 Any others? Can we prove/disprove?

$${n}={x}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{2}} =\mathrm{2}×\mathrm{2} \\ $$$$ \\ $$$$\mathrm{Any}\:\mathrm{others}?\:\mathrm{Can}\:\mathrm{we}\:\mathrm{prove}/\mathrm{disprove}? \\ $$

Answered by RasheedSindhi last updated on 27/Dec/15

log(n^x )=log(2n) x log n=log 2+log n x=((log 2+log n)/(log n)) x=1+((log 2)/(log n)) x=1+log_n 2 for n=2,x=2 let n=(√2) x=1+log_(√2) 2=1+2=3 ((√2))^3 =2((√2)) for n=^m (√2),x=m+1,m∈N (^m (√2))^(m+1) =2(^m (√2)) Verification: (^m (√2))^m .^m (√2)=2(^m (√2)) 2(^m (√2))=2(^m (√2))

$$\mathrm{log}\left(\mathrm{n}^{\mathrm{x}} \right)=\mathrm{log}\left(\mathrm{2n}\right) \\ $$$$\mathrm{x}\:\mathrm{log}\:\mathrm{n}=\mathrm{log}\:\mathrm{2}+\mathrm{log}\:\mathrm{n} \\ $$$$\:\:\:\:\:\:\mathrm{x}=\frac{\mathrm{log}\:\mathrm{2}+\mathrm{log}\:\mathrm{n}}{\mathrm{log}\:\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}=\mathrm{1}+\frac{\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{1}+\mathrm{log}_{\mathrm{n}} \mathrm{2} \\ $$$$\:{for}\:\:\:\mathrm{n}=\mathrm{2},\mathrm{x}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{let}\:\mathrm{n}=\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{1}+\mathrm{log}_{\sqrt{\mathrm{2}}} \mathrm{2}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{2}\left(\sqrt{\mathrm{2}}\right) \\ $$$${for}\:{n}=^{{m}} \sqrt{\mathrm{2}},{x}={m}+\mathrm{1},{m}\in\mathbb{N} \\ $$$$\:\:\:\left(^{{m}} \sqrt{\mathrm{2}}\right)^{{m}+\mathrm{1}} =\mathrm{2}\left(^{{m}} \sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{Verification}: \\ $$$$\:\:\:\:\left(^{{m}} \sqrt{\mathrm{2}}\right)^{{m}} .^{{m}} \sqrt{\mathrm{2}}=\mathrm{2}\left(^{{m}} \sqrt{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\mathrm{2}\left(^{{m}} \sqrt{\mathrm{2}}\right)=\mathrm{2}\left(^{{m}} \sqrt{\mathrm{2}}\right) \\ $$

Commented by Filup last updated on 27/Dec/15

What does this mean?

$$\mathrm{What}\:\mathrm{does}\:\mathrm{this}\:\mathrm{mean}? \\ $$

Commented by RasheedSindhi last updated on 27/Dec/15

For all n=2^(1/m) ,where m is any natural, There exist x=m+1 an integer such that (2^(1/m) )^(m+1) =2(2^(1/m) )

$$\mathrm{For}\:\mathrm{all}\:\mathrm{n}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{m}}} ,{where}\:{m}\:{is}\:{any} \\ $$$${natural},\:{There}\:{exist}\:{x}={m}+\mathrm{1} \\ $$$${an}\:{integer}\:{such}\:{that} \\ $$$$\:\:\:\left(\mathrm{2}^{\mathrm{1}/{m}} \right)^{{m}+\mathrm{1}} =\mathrm{2}\left(\mathrm{2}^{\mathrm{1}/{m}} \right) \\ $$

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