<s>[INST] How to show without calculator that $\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor$ By wolfram alpha, I get
$\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor=2996$.
How to prove that $\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor$ without calculator or wolfram alpha?
Thank in advances.
 [/INST] Prove that:
$$\log_{10}999^{999}+\log_{10}2<\log_{10}1000^{999}=2997$$
In other word:
$$\log_{10}2<\log_{10}1000^{999}-\log_{10}999^{999}=\log_{10}\left(\frac{1000}{999}\right)^{999}$$
so:
$$2<\left(\frac{1000}{999}\right)^{999}=\left(1+\frac{1}{999}\right)^{999}$$
It's true by Bernoulli's inequality.
Next we should prove $3 \cdot 999-1=2996<\log_{10}999^{999}$. It's equal:
$$3-\frac{1}{999}<\log_{10}999$$
or:
$$10^{3-\frac{1}{999}}=1000 \cdot 10^{-\frac{1}{999}}<999$$
$$10^{-\frac{1}{999}}<\frac{999}{1000}=\left(1-\frac{1}{1000}\right)$$
$$10^{-1}<\left(1-\frac{1}{1000}\right)^{999}$$
It's also true by Bernoulli's inequality.
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