Question

If $1,{\log }_9(3^{1-x}+2),{\log }_3(4\cdot 3^x-1)$ are in A.P., then $x$ is equal to

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

Answer 1

The correct option is B $1-{\log }_{3}4$

$1,{\log }_9(3^{1-x}+2),{\log }_3(4\cdot 3^x-1)$ are in A.P.
$\Rightarrow 2{\log }_9(3^{1-x}+2)=1+{\log }_3(4\cdot 3^x-1)$
$\Rightarrow \frac{2}{2}{\log }_3(3^{1-x}+2)={\log }_{3}3+{\log }_3(4\cdot 3^x-1)$
$\Rightarrow {\log }_3(3^{1-x}+2)={\log }_3\left(3\right(4\cdot 3^x-1\left)\right)$
$\Rightarrow 3^{1-x}+2=3(4\cdot 3^x-1)$
Put $3^x=t>0$
Then, $\frac{3}{t}+2=12t-3$
$\Rightarrow 12t^2-5t-3=0$
$\Rightarrow (4t-3)(3t+1)=0$
$\Rightarrow t=\frac{3}{4}(\because t>0)$
$\Rightarrow 3^x=\frac{3}{4}$
$\Rightarrow x={\log }_3\frac{3}{4}=1-{\log }_{3}4$