Question

If ${\log }_3(2x^2+6x-5)>1$, then the number of integral values of $x$ which does not satisfy the inequality is

Answer 1

For $\log$ to be defined,
$2x^2+6x-5>0$
$\Rightarrow x\in (-\infty ,\frac{-3-\sqrt{19}}{2})\cup (\frac{-3+\sqrt{19}}{2},\infty )\cdots \left(1\right)$

Now, ${\log }_3(2x^2+6x-5)>1$
$\Rightarrow 2x^2+6x-5>3^1\Rightarrow 2x^2+6x-8>0\Rightarrow x^2+3x-4>0\Rightarrow (x-1)(x+4)>0$
$\Rightarrow x\in (-\infty ,-4)\cup (1,\infty )\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$x\in (-\infty ,-4)\cup (1,\infty )$
So, the integers which does not satisfy the inequality are $-4,-3,-2,-1,0,1$