Answer
17
Work Step by Step
Let p represent the number of searches done then the number of comparisons will be-
Sequential search: p*100000
Sorting + Binary search: n*log(100000)+p*log(100000)
To find the case after which binary search becomes more efficient
$p*100000 \geq 100000*\log(100000)+p*\log(100000)$
=> $p*100000-p*\log(100000) \geq 100000*\log(100000)$
=> $p*(100000-\log(100000))\geq 100000*\log(100000)$
ignoring $\log(100000)$ in the left side,
=> $p*(100000)\geq 100000*\log(100000)$
=> $p\geq \log(100000)$
=> $p\geq 16.60$
Hence after 17 searches binary search will be better
