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Question Number 10040 by FilupSmith last updated on 21/Jan/17

$$\mathrm{Show}\mathrm{that}$$$$\lfloor {\log }_{10}\left(x\right)+1\rfloor \mathrm{give}\mathrm{the}\mathrm{number}\mathrm{of}$$$$\mathrm{digits}\mathrm{for}x\in \mathbb{Z}$$

Answered by mrW1 last updated on 21/Jan/17

$$letxbeanintegerwithndigits,then$$$${10}^{n-1}⩽x<{10}^n$$$${\log }_{10}{10}^{n-1}⩽{\log }_{10}\left(x\right)<{\log }_{10}{10}^n$$$$n-1⩽{\log }_{10}\left(x\right)<n$$$$n⩽{\log }_{10}\left(x\right)+1<n+1$$$$\Rightarrow \lfloor {\log }_{10}\left(x\right)+1\rfloor =n$$

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