If 1 , log 3 ( 3 1 - x + 2 ) , log 3 ( 4 · 3 x - 1 ) are in A.P., then x equals

Question

If $1, \log_{3} (3^{1 - x} + 2), \log_{3} (4 \cdot 3^{x} - 1)$ are in A.P., then $x$ equals

Smart Achievers · Accepted Answer

(2)

$As 1, \frac{1}{2} \log_{3} (3^{1 - x} + 2), \log_{3} (4 \cdot 3^{x} - 1) \in A.P.$

$\Rightarrow$ $\log_{3} (3^{1 - x} + 2) = \log_{3} (4 \cdot 3^{x} - 1) + 1$

$\Rightarrow$ $3^{1 - x} + 2 = (4 \cdot 3^{x} - 1) \times 3$ $[∵ \log_{3} 3 = 1]$

$\Rightarrow$ $3^{1 - x} + 2 = 12 \cdot 3^{x} - 3$

$\Rightarrow$ $3^{x} (3^{1 - x} + 2) = 12 \cdot 3^{2 x} - 3 \cdot 3^{x}$ $(Multiplying 3^{x} on both sides)$

$\Rightarrow$ $12 t^{2} - 5 t - 3 = 0$ $[where t = 3^{x}]$

$\Rightarrow$ $(3 t + 1) (4 t - 3) = 0$ $\Rightarrow$ $t = - \frac{1}{3}, t = \frac{3}{4}$

$\Rightarrow$ $3^{x} = - \frac{1}{3} which is not possible, and t = \frac{3}{4} \Rightarrow 3^{x} = \frac{3}{4}$

$\Rightarrow$ $x \log_{3} 3 = \log_{3} 3 - \log_{3} 4$ $(By taking logarithm at the base 3 both sides)$

$\Rightarrow$ $x = 1 - \log_{3} 4$

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