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Question Number 137473 by mr W last updated on 03/Apr/21

$$whatislarger?{99}^{100}or{100}^{99}?$$

Answered by MJS_new last updated on 03/Apr/21

$$n^{n+1}>{(n+1)}^n\mathrm{\forall }n⩾3$$

Commented by mr W last updated on 03/Apr/21

$$yessir.$$

Answered by mr W last updated on 03/Apr/21

$$f\left(x\right)=x^{\frac{1}{x}}=e^{\frac{\ln x}{x}}$$$$f^{\prime }\left(x\right)=x^{\frac{1}{x}}(\frac{1}{x^2}-\frac{\ln x}{x^2})=(1-\ln x)x^{\frac{1}{x}-2}$$$$forx<e:f^{\prime }\left(x\right)>0\Rightarrow f\left(x\right)isstrictlyincreasing$$$$forx>e:f^{\prime }\left(x\right)<0\Rightarrow f\left(x\right)isstrictlydecreasing$$$$thatmeansfor3⩽m<n:$$$$m^{\frac{1}{m}}>n^{\frac{1}{n}}$$$$\Rightarrow {\left(m^{\frac{1}{m}}\right)}^{mn}>{\left(n^{\frac{1}{n}}\right)}^{mn}$$$$\Rightarrow m^n>n^m$$$$so{99}^{100}>{100}^{99}$$

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