Prove that: `2(sqrt((log)_a4sqrt(a b)+(log)_b4sqrt(a b))-(log)_a4sqrt(2/b)+(log)_b4sqrt(a/b))dotsqrt((log)_a b)={2ifbgeqa >1 2^((log)_a bif1

Options

A. 1
B. 2
C. `2^(log_(a)b)`
D. ` 2^(log_(b)a)`

Answer

B

Explanation

We have
` E= 2^((sqrt(log_(a)root(4)(ab)+log_(b)root(4)(ab))-sqrt(log _(a)root(4)(b/a+log_(b)root(4)(a/b)))) sqrt(log_(a)b))`
` = 2^(1/2(sqrt(log_(a)ab+log_(b)ab-)sqrt(log_(a)b//a+log_(b)a//b))sqrt(log_(a)b))`
` = 2^(1/2(sqrt(2+log_(a)b+log_(b)a)-sqrt(log_(a)b+log_(b)a - 2))sqrt(log_(a)b))`
` = 2^(1/2(sqrt((log_(a)b)^(2)+2log_(a)b+1)-sqrt((log_(a)b)^(2)-2log_(a)b+1))`
` = 2^(1/2(sqrt((log_(a)b+1)^(2))-sqrt((log_(a)b-1)^(2)))`
` =2^(1/2(|log_(a)b+1|-|log_(a)b-1|)`
Case I:
` bgea gt1`
` rArr log_(a) b ge log_(a) a`
` rArr log_(a) b ge 1`
` rArrE=2^(1/2(log_(a)b+1-log_(a)b+1))=2`
Case II:
` 1 lt b lt a`
` rArr 0 lt log_(a) b lt log_(a) a`
` rArr 0 lt log_(a) b lt 1`
` rArr E = 2^(1/2(log_(a)b+1-1+log_(a)b))`
` = 2^(1//2.(2log_(a)b))`
` = 2 ^(log_(a)b)`