If-log-3-2-log-3-2-x-5-and-log-3-2-x-7-2-are-in-arithmetic-progression-then-find-the-value-of-x-

Question Number 258 by mchawla last updated on 25/Jan/15

If \log_{3} 2, \log_{3} (2^{x} - 5) and \log_{3} (2^{x} - \frac{7}{2}) are

in arithmetic progression then find

the value of x .

Answered by 123456 last updated on 17/Dec/14

\log_{3} 2, \log_{3} (2^{x} - 5), \log_{3} (2^{x} - \frac{7}{2})

since it was at pa then midle term is arithmetic mean of side term

{2 \log}_{3} (2^{x} - 5) = \log_{3} 2 + \log_{3} (2^{x} - \frac{7}{2})

{2 \log}_{3} (2^{x} - 5) = \log_{3} [2 (2^{x} - \frac{7}{2})]

\log_{3} {(2^{x} - 5)}^{2} = \log_{3} (2 \cdot 2^{x} - 7)

{(2^{x} - 5)}^{2} = 2 \cdot 2^{x} - 7

seting y = 2^{x}

{(y - 5)}^{2} = 2 y - 7

y^{2} - 10 y + 25 = 2 y - 7

y^{2} - 12 y + 32 = 0

Δ = {(- 12)}^{2} - 4 (1) (32)

= 144 - 128

= 16

y = \frac{- (- 12) \pm \sqrt{16}}{2 (1)}

= \frac{12 \pm 4}{2}

= 6 \pm 2

y_{1} = 4 \Rightarrow 2^{x} = 2^{2} \Leftrightarrow x = 2 + 2 π k i

\log_{3} 2, \log_{3} (- 1), \log_{3} \frac{1}{2}

\log_{3} 2, \log_{3} (- 1), - \log_{3} 2 (w t f)

y_{2} = 8 \Rightarrow 2^{x} = 2^{3} \Leftrightarrow x = 3 + 2 π k i

\log_{3} 2, \log_{3} 3, \log_{3} \frac{9}{2}

\log_{3} 2, 1, 2 - \log_{3} 2 (r = 1 - \log_{3} 2)

ignore blue part, k \in Z

then x = 3

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