How to show without calculator that $\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor$

Question

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.

Answer 1

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.