InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
`2^((sqrt(log_a(ab)^(1/4)+log_b(ab)^(1/4))-sqrt(log_a(b/a)^(1/4)+log_b(a/b)^(1/4)))sqrt(log_a(b))` =A. 1B. 2C. ` 2^(log_(a) b)`D. ` 2 ^(log_(b)a)` |
|
Answer» Correct Answer - C 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)` |
|