First slide

Arithmetic progression

Question

$If 1, \log_{3} \sqrt{(3^{1 - x} + 2)}, \log_{3} ({4.3}^{x} - 1) are in A.P then, x equals:$

Moderate

A

$\log_{3} 4$
B

$1 - \log_{3} 4$
C

$1 - \log_{4} 3$
D

$\log_{e} (\frac{4}{e})$

Solution

First of all logarithmic terms should have been defined,
$\begin{array}{l} \log_{3} \sqrt{(3^{1 - x} + 2)}, so 3^{1 - x} + 2 > 0 which is true \forall x \\ \log_{3} ({4.3}^{x} - 1), so \log_{3} ({4.3}^{x} - 1) thus 3^{x} > \frac{1}{4} \\ Now, terms are in A.P so, \\ 2 \log_{3} [\sqrt{(3^{1 - x} + 2)}] = 1 + \log_{3} (4 \cdot 3^{x} - 1) \\ = \log_{3} 3 + \log_{3} ({4.3}^{x} - 1) = \log_{3} [3 ({4.3}^{x} - 1)] \\ {(\sqrt{3^{1 - x} + 2})}^{2} = 3 ({4.3}^{x} - 1) \\ 3^{1 - x} + 2 = 3 ({4.3}^{x} - 1) = {12.3}^{x} - 3^{1} \\ \frac{3^{1}}{3^{x}} + 2 = 12 \cdot 3^{x} - 3 \\ Let t = 3^{x} \\ \frac{3}{t} + 2 = 12 t - 3, so 3 + 2 t = 12 t^{2} - 3 t \\ 12 t^{2} - 5 t + 3 = 0 \\ 12 t^{2} - 9 t + 4 t - 3 = 0 \\ 3 t (4 t - 3) + 1 (4 t - 3) = 0 \end{array}$

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