If a,b,c are distinct positive real numbers and a²b³c⁵ =2¹⁰,then 2a+3b+5c is always greater than

a.10

b.20

c.1

d.5

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).

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.

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+...x_n)/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)

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