Question

If $1,{\log }_3\sqrt{3^{1-x}+2},{\log }_3(4.3^3-1)$ 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)

Since $1,lo{g}_3\sqrt{3^{1-x}+2},lo{g}_3\left(4\cdot 3^x-1\right)$ are in $A.P$.
$\therefore 2lo{g}_3\sqrt{3^{1-x}+2}=1+lo{g}_3\left(4.3^x-1\right)$
$lo{g}_3\left(3^{1-x}+2\right)=lo{g}_{3}3+lo{g}_3\left(4.3^x-1\right)$
$\Rightarrow 3^{1-x}+2=3\left(4.3^x-1\right)$
$\Rightarrow 3.3^{-x}+2=12.3^x-3$
Put $3^x=y$
$\therefore \frac{3}{y}+2=12y-2$
$\Rightarrow 12y^2-4y-3=0$
$\Rightarrow \left(4y-3\right)\left(3y+1\right)=0$
$\Rightarrow y=\frac{3}{4},-\frac{1}{3}$
But $y=3^x$
$\therefore 3^x=\frac{3}{4}or-\frac{1}{3}$
But $3^x$ cannot be $-ve$
$\therefore 3^x=\frac{3}{4}$
$\Rightarrow xlo{g}_{3}3=lo{g}_3\left(\frac{3}{4}\right)$
$=lo{g}_{3}3-lo{g}_{3}4=1-lo{g}_{3}4$
$\Rightarrow x=1-lo{g}_{3}4$