Question : What is the number of real roots of $(\log x)^2- \lfloor\log x\rfloor-2=0$. $\lfloor\,\cdot\,\rfloor$ represents the greatest integer function less than or equal to $x$.

I know how to solve logarithm equation but due to greatest integer function I am unable to proceed further please help thanks.


Since $ [\log x] \leq \log x $

we have $(\log x)^2-\log x -2 \leq 0$

This is equivalent to $-1 \leq \log x \leq 2$

When $-1 \leq \log x \leq 0, [\log x ] =-1$ so that $\log x =\pm 1$ If we see that $\log x =1$ is not in the specified range. Hence $\log x =-1$ and $x =\frac{1}{10}$

When $0 \leq \log x < 1$ , $[\log x] =0$ so that $\log x =\pm \sqrt{2}$ None of these values in the range.

Similarly we can use $1 \leq \log x < 2$ this will give us $x =10^{\sqrt{3}}$

When $\log x =2$, $[\log x] =2$ and equation is satisfied. Thus $x =100$ is third real root.

  • $\begingroup$ we need to consider $$\log x-1<[\log x]$$ as well $\endgroup$ – lab bhattacharjee Feb 23 '14 at 18:38

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