Sequence And Series Question 63
Question: . If $ {\log_3}2,\ {\log_3}(2^{x}-5) $ and $ {\log_3}( 2^{x}-\frac{7}{2} ) $ are in A.P., then $ x $ is equal to
[IIT 1990]
Options:
A) $ 1,\ \frac{1}{2} $
B) $ 1,\ \frac{1}{3} $
C) $ 1,\ \frac{3}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ {\log_3}2,\ {\log_3}(2^{x}-5) $ and $ {\log_3}( 2^{x}-\frac{7}{2} ) $ are in A.P.
$ \Rightarrow $ $ 2{\log_3}(2^{x}-5)={\log_3}[ (2),( 2^{x}-\frac{7}{2} ) ] $
$ \Rightarrow $ $ {{(2^{x}-5)}^{2}}={2^{x+1}}-7 $
$ \Rightarrow $ $ 2^{2x}-12\ .\ 2^{x}-32=0 $
$ \Rightarrow $ $ x=2,\ 3 $ But $ x=2 $ does not hold, hence $ x=3 $ .
BETA
