Equations And Inequalities Question 42
Question: The number of real values of parameter k for which $ {{(log_{16}x)}^{2}}-{\log_{16}}x+{\log_{16}}k=0 $ will have exactly one solution is
Options:
A) 0
B) 2
C) 1
D) 4
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] The equation is $ {{(log_{16}x)}^{2}}-{\log_{16}}x+{\log_{16}}k=0 $ Clearly $ x>0 $ Solving the equation, we get $ {\log_{16}}x=\frac{1\pm \sqrt{1-4(log_{16}k)}}{2} $ For exactly one solution $ 1-4{\log_{16}}k=0 $
$ \Rightarrow k^{4}=16\Rightarrow k=\pm 2 $ [Taking real values] Now $ {\log_{16}}k $ is defined if $ k>0\Rightarrow k=2 $
BETA
