If $\log_{1/2}(x^2-5x+7)>0$ then exhaustive range of values of x is
Answer & explanation
Correct answer: option 2
$\log_{1/2}(x^2-5x+7)>0⇒x^2-5x+7≤(\frac{1}{2})0$
as $\frac{1}{2}<1$
So $x^2-5x+7<1⇒x^2-5x+6<0$
So $(x-2)(x-3)<0$
$⇒x∈(2, 3)$
If $\log_{1/2}(x^2-5x+7)>0$ then exhaustive range of values of x is
Correct answer: option 2
$\log_{1/2}(x^2-5x+7)>0⇒x^2-5x+7≤(\frac{1}{2})0$
as $\frac{1}{2}<1$
So $x^2-5x+7<1⇒x^2-5x+6<0$
So $(x-2)(x-3)<0$
$⇒x∈(2, 3)$