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This question already has an answer here:

Let $x$ and $y$ be positive numbers. Prove that $$\lfloor \log_{10}(xy)\rfloor \geq \lfloor \log_{10}(x)\rfloor+ \lfloor \log_{10}(y)\rfloor.$$

I thought about using the definition of floor functions but didn't see how to use that here.

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marked as duplicate by Winther, Alex Mathers, Shailesh, Daniel W. Farlow, user91500 Sep 22 '16 at 6:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Winther Yes, typo. $\endgroup$ – user19405892 Sep 21 '16 at 20:47
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Since $\log_{10}(xy)=\log_{10}(x)+\log_{10}(y)$, it's enough to prove that $\lfloor a+b\rfloor \geq \lfloor a\rfloor +\lfloor b\rfloor$ for all real numbers $a,b$.

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Hint:

$\bigl\lfloor\log_{10}x\bigr\rfloor=k\iff 10^k\le x<10^{k+1}$.

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