Prove that $$\Big(\frac{21}{20}\Big)^{100}>100.$$ I have tried proving that $$100\log\Big(\frac{21}{20}\Big)>2$$ but I was not able to evaluate it properly.
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Well, try to break down the logarithm, using the rules of logarithms, so that it will become something that a calculator can actually calculate. So instead of "to the power 100", you'll have "multiplied by 100". – Matti P. Apr 29 '20 at 09:45
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\begin{align}\Big({21\over20}\Big)^{100} &= (1+{1\over 20})^{100}\\ &={100\choose 0} +{100\choose 1} ({1\over 20})^{1} +{100\choose 2} ({1\over 20})^{2} +{100\choose 3} ({1\over 20})^{3} +... \\ &>1+5+12+20+24,5+23,5+18 =105 \end{align}
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I think you are on the right track. Using the basic rules for logartihms we get:
$\log(\frac{21^{100}}{20}) = 100\log(21) - \log(20) > 100 - 1 - \log(2) > 2$
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