Question

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

Answer 1

$2{\log }_{1/2}(x-1)\le \frac{1}{3}-\frac{1}{{\log }_{x^2-x}8}$
$-2{\log }_2(x-1)\le \frac{1}{3}-\frac{1}{{\log }_{x^2-x}8}[\because {\log }_{1/a}b=-{\log }_{a}b]$
$-2{\log }_2(x-1)+\frac{1}{{\log }_{x^2-x}8}\le \frac{1}{3}$
$-2{\log }_2(x-1)+{\log }_8(x^2-x)\le \frac{1}{3}[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$

$-2{\log }_2(x-1)+\frac{\log (x^2-x)}{3\log 2}\le \frac{1}{3}[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$

$-2{\log }_2(x-1)+\frac{1}{3}\left[{\log }_2\right(x-1)+{\log }_2(x\left)\right]\le \frac{1}{3}[\because \log (ab)=\log a+\log b]$

${\log }_2\left[\right(x-1{)}^{1/3}\times x^{1/3}\times (x-1{)}^{-2}]\le \frac{1}{3}[\because x\log a=\log a^m]$
${\log }_2\left[\right(x-1{)}^{-5/3}\times x^{1/3}]\le \frac{1}{3}$
${\log }_2\left[\right(x-1{)}^{-5}\times x]\le 1$

$\Rightarrow \left[\right(x-1{)}^{-5}\times x]\le 2$

$\Rightarrow x\le 2(x-1{)}^5$

If $x\le 0$, the L.H.S. is greater than R.H.S. and so those integers don't satisfy the inequality.

If both numbers $x$ and $x-1$ are greater than $1$ then inequality holds true.[Because $x^m>x^n,x>1\&m>n$

But RHS is also multiplied by $2$ so,

So, $x\ge 1\&x-1\ge 1$

$\Rightarrow x\ge 2$
Now, from $x=2$ to $x=10$, we get the inequality to be correct, hence the number of integers is $9$.