Sequence And Series Question 384
Question: If $ {\log_{10}}2,,{\log_{10}}(2^{x}-1) $ and $ {\log_{10}}(2^{x}+3) $ are three consecutive terms of an A.P, then the value of x is
Options:
A) 1
B) $ log_52 $
C) $ log_25 $
D) $ log_{10}5 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ {\log_{10}}2,,{\log_{10}}(2^{x}-1) $ and $ {\log_{10}}(2^{x}+3) $ are in A.P. Hence, common difference will be same.
$ \therefore {\log_{10}},(2^{x}-1)-{\log_{10}}2 $ $ =\log ,(2^{x}+3)-{\log_{10}}(2^{x}-1) $
$ \therefore ,{\log_{10}}( \frac{2^{x}-1}{2} )={\log_{10}}( \frac{2^{x}+3}{2^{x}-1} ) $
$ \Rightarrow ,\frac{2^{x}-1}{2}=\frac{2^{x}+3}{2^{x}-1} $ $ {{(2^{x}-1)}^{2}}=2(2^{x}+3) $ $ 2^{2x}-{2^{x+1}}+1={2^{x+1}}+6 $ $ 2^{2x}-{2^{x+2}}=5 $ Let $ 2^{x}=y, $ then $ y^{2}-4y-5=0 $ $ y^{2}-5y+y-5=0 $ $ y(y-5)+1(y-5)=0 $ $ y=-1,,y=5 $ Therefore, $ 2^{x}=5 $ $ x={\log_2}5. $
BETA
