Question

If $1$, ${\log }_9\left(3^{1-x}+2\right),{\log }_3\left({4.3}^x-1\right)$ are in $A.P.$ then $x$ equals ______________.

• A. ${\log }_{3}4$
• B. $1-{\log }_{3}4$
• C. $1-{\log }_{4}3$
• D. ${\log }_{4}3$

Answer 1

Answer: (B)

$1-{\log }_{3}4$

$1,{\log }_9(3^{1-x}+2),{\log }_3(4.3^x-1)$ are in A.P

$\Rightarrow 2{\log }_9(3^{1-x}+2)=1+{\log }_3(4.3^x-1)$

$\Rightarrow 2{\log }_{3^2}(3^{1-x}+2)=1+{\log }_3(4.3^x-1)$
$\Rightarrow 2\left(\frac{1}{2}{\log }_3(3^{1-x}+2)\right)=1+{\log }_3(4.3^x-1)$

$\Rightarrow {\log }_3(3^{1-x}+2)={\log }_{3}3+{\log }_3(4.3^x-1)$

$\Rightarrow {\log }_3(3^{1-x}+2)={\log }_{3}3(4.3^x-1)$

$\Rightarrow 3^{1-x}+2=3(4.3^x-1)$

$\Rightarrow 3\frac{1}{3^x}+2=12.3^x-3$

$\Rightarrow 3+23^x=12.3^{2x}-33^x$

$\Rightarrow 12.3^{2x}-53^x-3=0$

Let $\alpha =3^x\Rightarrow {\alpha }^2=3^{2x}$

$\Rightarrow 12{\alpha }^2-5\alpha -3=0$

$\Rightarrow 12{\alpha }^2-9\alpha +4\alpha -3=0$

$\Rightarrow 3\alpha (4\alpha -3)+1(4\alpha -3)=0$

$\Rightarrow (4\alpha -3)(3\alpha +1)=0$

$\Rightarrow 4\alpha -3=0,3\alpha +1=0$

$\Rightarrow 4\alpha =3,3\alpha =-1$

$\Rightarrow \alpha =\frac{3}{4},\alpha =\frac{-1}{3}$

$\Rightarrow 3^x=\frac{3}{4},3^x=\frac{-1}{3}$ where $\alpha =3^x$

$\Rightarrow 3^x=\frac{-1}{3}$ is not possible

Hence $3^x=\frac{3}{4}$

$\Rightarrow {\log }_3{3}^x={\log }_3\frac{3}{4}$

$\Rightarrow x{\log }_{3}3={\log }_{3}3-{\log }_{3}4$

$\therefore x=1-{\log }_{3}4$