If $1, {log}_{9} (3^{1 - x} + 2), {log}_{3} [{4.3}^{x} - 1]$ are in $A P$ , then $x$ equals

{log}_{3} 4

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1 - {log}_{3} 4

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1 - {log}_{4} 3

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{log}_{4} 3

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Solution

The correct option is C $1 - {log}_{3} 4$
$\begin{matrix} G i v e n, 1, {log}_{9} (3^{1 - x} + 2), {log}_{3} [{4 \cdot 3}^{x} - 1] a r e i n A . P . \Rightarrow 2 {log}_{9} (3^{1 - x} + 2) = 1 + {log}_{3} [{4 \cdot 3}^{x} - 1] \Rightarrow {log}_{3} (3^{1 - x} + 2) = {log}_{_{3}} 3 + {log}_{3} [{4 \cdot 3}^{x} - 1] = {log}_{3} (3 ({4 \cdot 3}^{x} - 1)) \Rightarrow 3^{1 - x} + 2 = 3 ({4 \cdot 3}^{x} - 1) \Rightarrow 3^{- x} + \frac{2}{3} = 4 \cdot 3^{x} - 1 \end{matrix}$
$\Rightarrow \frac{1}{3^{x}} + \frac{2}{3} + 1 = 4 \cdot 3^{x}$
$\Rightarrow \frac{1}{3^{x}} + \frac{5}{3} = 4 \cdot 3^{x}$

$L e t 3^{x} = t$

$\Rightarrow \frac{1}{t} + \frac{5}{3} = 4 t$

$\Rightarrow 12 t^{2} - 5 t - 3 = 0$
$\Rightarrow 12 t^{2} + 4 t - 9 t - 3 = 0$
$\Rightarrow 4 t (3 t + 1) - 3 (3 t + 1) = 0$

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

$t = \frac{3}{4} O R t \neq \frac{- 1}{3}$ ......as $log$ can't take negative values

$3^{x} = \frac{3}{4}$
taking $log$ on both sides
$log (3^{x}) = log \frac{3}{4}$

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

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