Question

The solution set of inequality ${\log }_5(x-3)+\frac{1}{2}{\log }_{5}3<\frac{1}{2}{\log }_5(2x^2-6x+7)$ is

• A. $(2,10)$
• B. $(3,10)$
• C. $(3,\infty )$
• D. $(4,10)$

Answer 1

The correct option is B $(3,10)$

${\log }_5(x-3)+\frac{1}{2}{\log }_{5}3<\frac{1}{2}{\log }_5(2x^2-6x+7)$
$\log$ function is defined when
$\left(i\right)x>3$
$\left(ii\right)2x^2-6x+7>0$
As $D<0,$ so $2x^2-6x+7>0\forall x\in R$

Now, $2{\log }_5(x-3)+{\log }_{5}3<{\log }_5(2x^2-6x+7)$
$\Rightarrow {\log }_{5}3(x-3{)}^2<{\log }_5(2x^2-6x+7)$
$\Rightarrow 3(x-3{)}^2<(2x^2-6x+7)$
$\Rightarrow 3x^2-18x+27<2x^2-6x+7$
$\Rightarrow x^2-12x+20<0$
$\Rightarrow (x-2)(x-10)<0$
$\Rightarrow x\in (2,10)...\left(iii\right)$

From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$x\in (3,10)$