First slide

Arithmetic progression

Question

If $\log_{10} 2, \log_{10} (2^{x} + 1)$ and $\log_{10} (2^{x} + 3)$ are in A.P., then

Moderate

A

x = 0
B

x = 1
C

$x = \frac{1}{2} \log_{10} 2$
D

$x = \frac{1}{2} \log_{2} 5$

Solution

$2, 2^{x} + 1, 2^{x} + 3$ are in G.P.
$\Rightarrow {(2^{x} + 1)}^{2} = 2 (2^{x} + 3) \Rightarrow 2^{2 x} = 5$
$\Rightarrow x = \frac{1}{2} \log_{2} 5$

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