First slide

Arithmetic progression

Question

If $\log_{2} (5 {.2}^{x} + 1), \log_{4} (2^{1 - x} + 1)$ and 1 are in A.P, then x equals

Moderate

A

$\log_{2} 5$
B

$1 - \log_{2} 5$
C

$\log_{5} 2$
D

None of these

Solution

The given numbers are in A.P.
$\begin{aligned} ∴ 2 \log_{4} (2^{1 - x} + 1) = \log_{2} (5 {.2}^{x} + 1) + 1 \\ \Rightarrow & 2 \log_{2^{2}} (\frac{2}{2^{x}} + 1) = \log_{2} (5 {.2}^{x} + 1) + \log_{2} 2 \\ \Rightarrow & \frac{2}{2} \log_{2^{}} (\frac{2}{2^{x}} + 1) = \log_{2} (5 {.2}^{x} + 1) 2 \\ \Rightarrow & \log_{2} (\frac{2}{2^{x}} + 1) = \log_{2} (10 {.2}^{x} + 2) \\ \Rightarrow & \frac{2}{2^{x}} + 1 = 10 {.2}^{x} + 2 \\ \Rightarrow & \frac{2}{y} + 1 = 10 y + 2, where 2^{x} = y \\ ∴ & 10 y^{2} + y - 2 = 0 \\ \Rightarrow & (5 y - 2) (2 y + 1) = 0 \\ \Rightarrow & y = 2 / 5 \\ \Rightarrow & 2^{x} = 2 / 5 \Rightarrow x = \log_{2} (2 / 5) = \log_{2} 2 - \log_{2} 5 = 1 - \log_{2} 5 \end{aligned}$

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