Equations And Inequalities Question 26
Question: If the equation $ 2^{x}+4^{y}=2^{y}+4^{x} $ is solved for y in terms of x where $ x<0 $ , then the sum of the solutions is
Options:
A) $ x{\log_2}(1-2^{x}) $
B) $ x+{\log_2}(1-2^{x}) $
C) $ {\log_2}(1-2^{x}) $
D) $ x{\log_2}(2^{x}+1) $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] $ 2^{2y}-2^{y}+2^{x}(1-2^{x})=0 $ put $ 2^{y}=t $ $ t^{2}-t+2^{x}(1-2^{x})=0 $ where $ t_1={2^{y_1}} $ and $ t_2={2^{y_2}} $ $ t_1t_2=2^{x}(1-2^{x}) $ $ {2^{y_1+y_2}}=2^{x}(1-2^{x}) $ $ y_1+y_2=x+{\log_2}(1-2^{x}) $
BETA
