Positive integers ' a ' and 'b' satisfy the condition log 2( log 2 a ( log 2 b (21000)))=0. Find the sum of all possible values of ' a '.

Question

Positive integers ' a ' and 'b' satisfy the condition log 2( log 2 a ( log 2 b (21000)))=0. Find the sum of all possible values of ' a '. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

log 2 a ( log 2 b (21000))=1 log 2b(21000)=2a ⇒ 21000=(2b)2a=2b ⋅ 2a Hence, b ⋅ 2 a =1000

Q. Positive integers ' $a$ ' and 'b' satisfy the condition $lo g_{2} (lo g_{2^{a}} (lo g_{2^{b}} (2^{1000}))) = 0$ . Find the sum of all possible values of ' $a$ '.

$lo g_{2^{a}} (lo g_{2^{b}} (2^{1000})) = 1$
$lo g_{2^{b}} (2^{1000}) = 2^{a} \Rightarrow 2^{1000} = (2^{b})^{2^{a}} = 2^{b \cdot 2^{a}}$
Hence, $b \cdot 2^{a} = 1000$

Q. Positive integers ' a ' and 'b' satisfy the condition log2​(log2a​(log2b​(21000)))=0. Find the sum of all possible values of ' a '.

log2a​(log2b​(21000))=1 log2b​(21000)=2a⇒21000=(2b)2a=2b⋅2a Hence, b⋅2a=1000

Q. Positive integers ' $a$ ' and 'b' satisfy the condition $lo g_{2} (lo g_{2^{a}} (lo g_{2^{b}} (2^{1000}))) = 0$ . Find the sum of all possible values of ' $a$ '.

$lo g_{2^{a}} (lo g_{2^{b}} (2^{1000})) = 1$
$lo g_{2^{b}} (2^{1000}) = 2^{a} \Rightarrow 2^{1000} = (2^{b})^{2^{a}} = 2^{b \cdot 2^{a}}$
Hence, $b \cdot 2^{a} = 1000$