1guys try out .....atleast once attempt.....its very easy
if x_{n} is a monotonic increasing sequence withx_{1} > 1 i.e, 1<x_{1} < x_{2} <x_{3} < x_{4}<.............<x_{n} <x_{n+1} <..........
then find value of log_{x_{1}} } log_{x_{2}} } log_{x_{3}} } ................ log_{x_{n}} }x_{n}^{x(n-1)}x(n-2).. x_{2} ^{x_{1}}
-
UP 0 DOWN 0 0 6
6 Answers
1yup but i need a explaination bcoz some how i m tugged in b/w my solution
1k
use these 2 properties
1.log ab=b . loga
2. logaa=1
step by step all the powers will come out leaving behind an expression like prop 2 inthe end thus whole expression will b conversted to 1 step by step
1yup thanxx ........ i have succeded in solving the question .......
in the end it wuld b
log_{x_{1}} x_{1} x_{2} log_{x_{2}} x_{2}
log_{x_{1}} x_{1} = 1