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I have the following equation (log base 10):

$$\frac{x}{10^{\lfloor \log x/10 \rfloor}}$$ how can I show what the maximum value of this expression can be? i.e. $\frac{x}{10^{\lfloor \log x/10 \rfloor}} < y$.

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  • $\begingroup$ Is your equation meant to be $$\frac{x}{10 \lfloor \log x \rfloor}$$ $\endgroup$ – Zain Patel Jul 12 '16 at 15:11
  • $\begingroup$ ah no by ** i meant to say to the power, how do you guys write these equations? $\endgroup$ – Har Jul 12 '16 at 15:11
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    $\begingroup$ Edited it in for you. $\endgroup$ – Zain Patel Jul 12 '16 at 15:12
  • $\begingroup$ @barakmanos could you please show your steps on how you managed to get there? $\endgroup$ – Har Jul 12 '16 at 15:21
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For your original question:

Let $n$ denote the number of decimal digits in the integer part of $x$:

  • $x = d_{1} \dots d_{n}$
  • $\lfloor\log{x}\rfloor = n-1$
  • $10^{\lfloor\log{x}\rfloor} = 1\underbrace{0\dots0}_{n-1\text{ times}}$

Therefore:

$$\frac{x}{10^{\lfloor\log{x}\rfloor}} = \frac{d_{1} \dots d_{n}}{1\underbrace{0\dots0}_{n-1\text{ times}}}<10^1$$


For your updated question:

Let $n$ denote the number of decimal digits in the integer part of $x$:

  • $x = d_{1} \dots d_{n}$
  • $\lfloor\log{x}\rfloor = n-1$
  • $\lfloor\log{x/10}\rfloor = n-2$
  • $10^{\lfloor\log{x/10}\rfloor} = 1\underbrace{0\dots0}_{n-2\text{ times}}$

Therefore:

$$\frac{x}{10^{\lfloor\log{x/10}\rfloor}} = \frac{d_{1} \dots d_{n}}{1\underbrace{0\dots0}_{n-2\text{ times}}}<10^2$$

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