Sequence-And-Series Question 651
Question: If $ {\log_{10}}2,,{\log_{10}}(2^{x}-1),log_{10}(2^{x}+3) $ are three consecutive terms of an AP, then which one of the following is correct?
Options:
A) $ x=0 $
B) $ x=1 $
C) $ x={\log_2}5 $
D) $ x={\log_5}2 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ {\log_{10}}2,log_{10}( 2^{x}-1 ) $ and $ {\log_{10}}( 2^{x}+3 ) $ are in A.P
$ \therefore 2{\log_{10}}( 2^{x}-1 ) $ $ ={\log_{10}}^{2}+{\log_{10}}( 2^{x}+3 ) $
$ \Rightarrow {\log_{10}}{{( 2^{x}-1 )}^{2}}={\log_{10}}2( 2^{x}+3 ) $
$ \Rightarrow ,2^{2x}+1-{2^{x+1}}={{2.2}^{x}}+6 $
$ \Rightarrow ,a^{2}+1-2a=2a+6 $ where $ a=2^{x} $ .
$ \Rightarrow ,a^{2}-4a-5=0 $
$ \Rightarrow ,a=5 $ or $ a=-1 $ $ 2^{x}=5\Rightarrow \log 2=\log 5 $
$ \Rightarrow x=\frac{\log 5}{\log 2}\Rightarrow x={\log_2}5 $
BETA
