Question

If $x\in R$, ${\left(\log \right\{x\left\}\right)}^2-3\log \left[x\right]+2=0$ and $\frac{1-2x}{3-|x-1|}=1,$ where $\{.\}$ is the fractional part function and $[.]$ is the greatest integer function, then the number of non-integral value(s) of $x$ is

Answer 1

There are two simultaneous equations. So, we will try to solve the easier one first.
$\frac{1-2x}{3-|x-1|}=1$
$\Rightarrow 1-2x=3-|x-1|$
$\Rightarrow |x-1|=2(1+x)$

If $x\ge 1$, we get
$x-1=2(1+x)\Rightarrow x=-3$
Not possible

If $x<1$, we get
$1-x=2(1+x)\Rightarrow x=-\frac{1}{3}$

Now, putting $x=-\frac{1}{3}$, we get
${\left(\log \right\{x\left\}\right)}^2-3\log \left[x\right]+2=0\Rightarrow {\left(\log \{-\frac{1}{3}\}\right)}^2-3\log [-\frac{1}{3}]+2=0$
Which is not defined as $[-\frac{1}{3}]=0$

Hence, no solution exists.