The number N=( operatornameantilog5( log √5(14))) ⋅( log 9( operatornameantilog3(2-1))), then √ N is relatively prime with

Question

The number N=( operatornameantilog5( log √5(14))) ⋅( log 9( operatornameantilog3(2-1))), then √ N is relatively prime with (A) 30 (B) 42 (C) 70 (D) 105

Tardigrade Admin · Accepted Answer

N =(5 log √5 14)( log 9 32-1)=(14)2 × (2-1/2)=49 ⇒ √ N =7

Q. The number $N = (antilog_{5} (lo g_{5} (14))) \cdot (lo g_{9} (antilog_{3} (2^{- 1})))$ , then $N$ is relatively prime with

30

42

70

105

$N = (5^{l o g_{5} 14}) (lo g_{9} 3^{2^{- 1}}) = (14)^{2} \times \frac{2 ^{- 1}}{2} = 49$
$\Rightarrow N = 7$

Q. The number N=(antilog5​(log5​​(14)))⋅(log9​(antilog3​(2−1))), then N​ is relatively prime with

30

42

70

105

N=(5log5​​14)(log9​32−1)=(14)2×22−1​=49 ⇒N​=7

Q. The number $N = (antilog_{5} (lo g_{5} (14))) \cdot (lo g_{9} (antilog_{3} (2^{- 1})))$ , then $N$ is relatively prime with

$N = (5^{l o g_{5} 14}) (lo g_{9} 3^{2^{- 1}}) = (14)^{2} \times \frac{2 ^{- 1}}{2} = 49$
$\Rightarrow N = 7$