What is the number of real roots of $(\log x)^2- \lfloor\log x\rfloor-2=0$ $\lfloor\,\cdot\,\rfloor$ represents the greatest integer 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. 
