A) 0
B) 2
C) 1
D) 4
Correct Answer: C
Solution :
[c] The equation is \[{{(lo{{g}_{16}}x)}^{2}}-{{\log }_{16}}x+{{\log }_{16}}k=0\] Clearly \[x>0\] Solving the equation, we get \[{{\log }_{16}}x=\frac{1\pm \sqrt{1-4(lo{{g}_{16}}k)}}{2}\] 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 \[k>0\Rightarrow k=2\]
You need to login to perform this action.
You will be redirected in
3 sec
