Question

Number of integers less than or equal to $10$ satisfying the inequality ${\log }_{1/2}(x-1)\le \frac{1}{3}-\frac{1}{{\log }_{x^2-x}8}$ is

Answer 1

Given:
${\log }_{1/2}(x-1)\le \frac{1}{3}-\frac{1}{{\log }_{x^2-x}8}$
For logarithm to be defined:
$I.x-1>0\Rightarrow x>1,II.x^2-x>0\&x^2-x\ne 1\Rightarrow x(x-1)>0\Rightarrow x\in (-\infty ,0)\cup (1,\infty )\&x^2-x\ne 1\Rightarrow x^2-x-1\ne 0\Rightarrow x\ne \frac{1\pm \sqrt{5}}{2}$
$\therefore x\in (1,\infty )-\left\{\frac{1+\sqrt{5}}{2}\right\}$

Now, Solving the inequality, we get:
${\log }_{1/2}(x-1)\le \frac{1}{3}-\frac{1}{{\log }_{x^2-x}8}\Rightarrow -{\log }_2(x-1)\le \frac{1}{3}-\frac{{\log }_2(x^2-x)}{3}\Rightarrow -3{\log }_2(x-1)\le 1-{\log }_2(x^2-x)\Rightarrow {\log }_2(x^2-x)-{\log }_2(x-1{)}^3\le 1\Rightarrow {\log }_2\frac{(x^2-x)}{(x-1{)}^3}\le 1$
$\Rightarrow {\log }_2\frac{x}{(x-1{)}^2}\le 1[\because x>1]$
As the base of the log is greater than $1$, so
$\Rightarrow \frac{x}{(x-1{)}^2}\le 2\Rightarrow x\le 2x^2-4x+2\Rightarrow 2x^2-5x+2\ge 0\Rightarrow (2x-1)(x-2)\ge 0\Rightarrow x\le \frac{1}{2}\text{or}x\ge 2$
Therefore, the required solution is $x\ge 2$
Hence, the required number of intergral solution is $9.$