Question

Find the sum of all integers satisfying the inequalities ${\log }_5\left(x-3\right)+\frac{1}{2}{\log }_{5}3<\frac{1}{2}{\log }_5\left(2x^2-6x+7\right)$ and ${\log }_{3}x+{\log }_{\sqrt{3}}x+{\log }_{\frac{1}{3}}x<6.$

• A. $42$
• B. $45$
• C. $30$
• D. $39$

Answer 1

Answer: (A)

$42$

$lo{g}_5(x-3)+{\log }_5{3}^{1/2}\le \frac{1}{2}{\log }_5(2x^2-6x+7)$

${\log }_5(x-3)\left(\sqrt{3}\right)<{\log }_5(2x^2-6x+7{)}^{1/2}$

$3(x^2+9-6x)<2x^2-6x+7$

$3x^2+27-18x<2x^2-6x+7$

$x^2-12x+20<0$

$x^2-12x-2x+20<0$

$x(x-10)-2(x-10)<0$

$(x-10)(x-2)<0\Rightarrow x\in (2,10)$

${\log }_{3}x+2{\log }_{3}x-{\log }_{3}x<6$

$2{\log }_3{x}^2<6$

$x^2<3^6$

$x<3^{3}x<27$

$=3,4,5,6,7,8,9$

$=\frac{7\left[12\right]}{2}6=42$ Ans $\left(A\right)$