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 $1, log_{3}\sqrt{3^{1-x} +2}, log_{3}\left(4\cdot3^{x}-1\right)$ are in $ A.P$.
$ \therefore 2 log_{3}\sqrt{3^{1-x}+2} = 1+log_{3} \left(4.3^{x}-1\right)$
$log_{3}\left(3^{1-x} +2\right) = log_{3}3+log_{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 x log_{3} 3 = log_{3}\left(\frac{3}{4}\right) $
$= log_{3}3 -log_{3}4 = 1-log_{3} 4$
$ \Rightarrow x=1-log_{3}4 $