Question

Number of integral values of x satisfying ${\left(\frac{3}{4}\right)}^{6x+10-x^2}<\left(\frac{27}{64}\right)$ is ___.

Answer 1

$6x+10–x^2>3$

$x^2–6x–7<0\Rightarrow (x-7)(x+1)<0$

$-1<x<7\Rightarrow x=0,1,\cdots \cdots \cdots \cdots \cdots ,6$