Question-7407

Question Number 7407 by Tawakalitu. last updated on 27/Aug/16

Commented by sou1618 last updated on 27/Aug/16

a n o t h e r s o l u t i o n

999^{2} < 1024^{2} = 2^{10 \times 2} = 2^{20}

s o 999^{2} < 2^{20} < 2^{999}

Commented by Tawakalitu. last updated on 27/Aug/16

T h a n k s s o m u c h

Answered by sandy_suhendra last updated on 28/Aug/16

l o g 2^{999} = 999 \times l o g 2 = 999 \times 0.3 = 299.7

l o g 999^{2} = 2 \times l o g 999 = 2 \times 2.99 = 5.98

l o g 2^{999} > l o g 999^{2} s o 2^{999} > 999^{2}

Commented by Tawakalitu. last updated on 27/Aug/16

T h a n k s s o m u c h .

Answered by FilupSmith last updated on 28/Aug/16

2^{999} \overset{?}{>} 999^{2}

let 999 = 2^{k}

k = \frac{\log 999}{\log 2}

∴ 2^{999} > 2^{2 \times \frac{\log 999}{\log 2}}

999 > 2 \times \frac{\log 999}{\log 2}

499.5 > \frac{3 \log (3) + \log (37)}{\log (2)}

499.5 > \log_{a} (t)

a^{499.5} > t \Rightarrow 2^{499.5} > 999 (a = 2, t = 999)

true

∴ LHS > RHS

∴ 2^{999} > 999^{2}

Commented by FilupSmith last updated on 28/Aug/16

my attempt at a proof without approximating

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