# Lowest bound on logarthmic equation with floor

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. $\frac{x}{10^{\lfloor \log x/10 \rfloor}} < y$.

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

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:

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

Let $n$ denote the number of decimal digits in the integer part of $x$:
• $x = d_{1} \dots d_{n}$
• $\lfloor\log{x}\rfloor = n-1$
• $\lfloor\log{x/10}\rfloor = n-2$
• $10^{\lfloor\log{x/10}\rfloor} = 1\underbrace{0\dots0}_{n-2\text{ times}}$
$$\frac{x}{10^{\lfloor\log{x/10}\rfloor}} = \frac{d_{1} \dots d_{n}}{1\underbrace{0\dots0}_{n-2\text{ times}}}<10^2$$