Question

The values of $x$ satisfying the inequality $4\left(9^x\right)-19\left(3^x\right)+12\le 0,$ lie in the interval $[A,B].$ Then

• A. $A={\log }_3\frac{3}{4}$
• B. $A={\log }_{1/3}\frac{4}{3}$
• C. $B={\log }_{3}4$
• D. $B={\log }_{4}3$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $A={\log }_3\frac{3}{4}$
B $A={\log }_{1/3}\frac{4}{3}$
C $B={\log }_{3}4$

$4\left(9^x\right)-19\left(3^x\right)+12\le 0$
Let $y=3^x$, we get
$4y^2-19y+12\le 0\Rightarrow (y-4)(4y-3)\le 0\Rightarrow y\in [\frac{3}{4},4]\Rightarrow 3^x\in [\frac{3}{4},4]\therefore x\in [{\log }_3\frac{3}{4},{\log }_{3}4]$