Sequence And Series Question 182
Question: If $ 1,{\log_9}({3^{1-x}}+2),{\log_3}({{4.3}^{x}}-1) $ are in A.P. then x equals
[AIEEE 2002]
Options:
A) $ {\log_3}4 $
B) $ 1-{\log_3}4 $
C) $ 1-{\log_4}3 $
D) $ {\log_4}3 $
Show Answer
Answer:
Correct Answer: B
Solution:
The given number are in A.P.
$ \therefore 2{\log_9}({3^{1-x}}+2)={\log_3}({{4.3}^{x}}-1)+1 $
Þ $ 2{\log_{3^{2}}}({3^{1-x}}+2)={\log_3}({{4.3}^{x}}-1)+{\log_3}3 $
Þ $ \frac{2}{2}{\log_3}({3^{1-x}}+2)={\log_3}[3({{4.3}^{x}}-1)] $
Þ $ {3^{1-x}}+2=3,({{4.3}^{x}}-1) $
Þ $ \frac{3}{y}+2=12y-3, $ where $ y=3^{x} $
Þ $ 12y^{2}-5y-3=0 $ $ y=\frac{-1}{3} $ or $ \frac{3}{4}\Rightarrow 3^{x}=\frac{-1}{3} $ or $ 3^{x}=\frac{3}{4} $ $ x={\log_3},(3/4) $
$ \Rightarrow x=1-{\log_3}4 $ .
BETA
