Last Activity: 13 Years ago
Hi Menka,
Us AM-GM Ineguality on 2As,3Bs, and 5Cs.
So (2A+3B+5C)/10 > [A2B3C5]1/10
Hence, 2A+3B+5C > 20.
Hope it helps.
Regards,
Ashwin (IIT Madras).
Last Activity: 13 Years ago
Let a,b,c Be three different real no. such that a2b3c5 =210
Putting log2 on both sides log2 (a2b3c5) =log2 210
= log2 (a2)+ log2(b3)+ log2(c5) =log2 (210)
= 2 log2 (a) + 3 log2(b)+ 5 log2(c) =10
= As per the general rule x > log2(x) ( for all positive real no. this rule is followed), then 2 (a) + 3 (b)+ 5 (c) > 10 as a> log2(a) , b> log2(b) , c> log2(c)
So (a.) must be the correct answer.
Last Activity: 13 Years ago
Sorry , option (b) is correct according to the AM-GM inequality for positive no.
2A+3B+5C = (A+A) +(B+B+B)+(C+C+C+C+C)
According to the AM-GM inequalty,
(x1+x2+x3+x4+...xn)/n > (x1x2x3...xn)1/n
So (A+A) +(B+B+B)+(C+C+C+C+C) consist of 10 terms thus
((A+A) +(B+B+B)+(C+C+C+C+C))/10 > ((A*A) *(B*B*B)+(C*C*C*C*C))1/10
So (2A+3B+5C)/10 > ((A2) *(B3)+(C5))1/10
thus (2A+3B+5C) /10 > (210)1/10
(2A+3B+5C) > (20) -----(Ans)
Last Activity: 12 Years ago
PLZZZZZZZZZZZZZZZZZZ APPROVE THE ANSWER IF IT HELPED YOU BY CLICKING THE YES BUTTON.......................................
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Get your questions answered by the expert for free
Last Activity: 1 Year ago
Last Activity: 1 Year ago
Last Activity: 2 Years ago