closed by
625 views
0 votes
0 votes
closed with the note: solved

THEY GIVE OPTIONS ARE 

  1. 0
  2. 1
  3. 2
  4. 3

WHAT I DO WAS I PUT P=8,Q=4,R=2 AND X=16 AND THEN GET RESULT ACCORDING TO THAT AND MY RESULT WAS CLOSE TO 1  BUT THEY ARE GIVING ANSWER 0 BY SOME THEOREM WHICH I DONT KNOW CORRECT OR NOT

anyones result 0 plzz prove that

closed by

1 Answer

1 votes
1 votes
$\frac{1}{Log_({p/q}) x} + \frac{1}{Log_({q/r}) x} +\frac{1}{Log_({r/p}) x}$

It is known that :-

${Log_{a} x}$ can be written as $\frac{Log_{x}x}{Log_{x} a}$.

Similarly , the above equation can be written as :-

${Log_{x} \frac{p}{q}}$ + ${Log_{x} \frac{q}{r}}$+${Log_{x} \frac{r}{p}}$ = ${Log_{x} (\frac{p}{q}*\frac{q}{r}*\frac{r}{p}})$ = $Log_{x}1$ = 0

Related questions

0 votes
0 votes
0 answers
1
superask asked Jan 30, 2017
304 views
0 votes
0 votes
0 answers
3
pream sagar asked Jan 23, 2019
502 views
0 votes
0 votes
0 answers
4
pream sagar asked Jan 23, 2019
627 views
Consider the system which has virtual address of 36 bits and physical address of 30 bits and page size of 8KB, page table entry contain 1 valid bit, 2 protection bit and ...