Binomial Theorem And Its Simple Applications Question 80
Question: The ninth term in the expansion of $ {{{ {3^{{\log_3}\sqrt{{25^{x-1}}+7}}}+{3^{-1/8{\log_3}( {5^{x-1}}+1 )}} }}^{10}} $ is equal to 180, then x is
Options:
A) A prime number
B) An irrational number
C) Has non-zero fractional part
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] We have, $ {{{ {3^{{\log_3}\sqrt{{25^{x-1}}+7}}}+{3^{-1/8{\log_3}( {5^{x-1}}+1 )}} }}^{10}} $
$ ={{[ \sqrt{{25^{x-1}}+7}+{{( {5^{x-1}}+1 )}^{-1/8}} ]}^{10}} $
(since $ {a^{{\log_{a}}N}}=N $ )
Here, $ T_9=180 $
$ \Rightarrow {{,}^{10}}C_8{{{ \sqrt{{25^{x-1}}+7} }}^{10-8}}{{{ {{( {5^{x-1}}+1 )}^{-1/8}} }}^{8}}=180 $
$ \Rightarrow {{,}^{10}}C_8( {25^{x-1}}+7 ){{( {5^{x-1}}+1 )}^{-1}}=180 $
$ \Rightarrow 45\frac{( {25^{x-1}}+7 )}{{5^{x-1}}+1}=180\Rightarrow \frac{{25^{x-1}}+7}{{5^{x-1}}+1}=4 $
$ \Rightarrow \frac{y^{2}+7}{y+1}=4, $ where $ y={5^{x-1}}\Rightarrow y^{2}-4y+3=0 $
$ \Rightarrow y=3,1\Rightarrow {5^{x-1}}=3 $ or $ {5^{x-1}}=1 $
$ \Rightarrow 5^{x}=15 $ or $ 5^{x}=5\Rightarrow x={\log_5}15 $ or $ x=1 $
BETA
