Where Quant meets Logic
Let,log(9) p = log(12) q = log(16) p+q = kHence,[log p]/2 = k log3 ------1[log(p+q)]/2 = k log4-------2log q = k log12--------3Adding 1 and 2-[log (p^2 + pq) ]/ 2 = k log12p^2 + pq = q^21+ q/p = (q/p)^2x^2 - x - 1 = 0(x-1/2)^2 = 5/4x = [1+-sqrt(5)]/2Since x=q/p>0 x= [1+sqrt5]/2
Let the three expressions be equal to k.Now it's easy 9^k+12^k=16^kWe can easily see that 2 satisfies.So q/p=144/81=16/9
Oops Monday morning blues ..Just ignore my above post:O
Ah, taking 9^k+12^k=16^kThen 1+(4/3)^k=(4/3)^2kNow the equation is t^2-t-1=0 where t=(4/3)^know t>0 so t=(1+sqrt(5))/2..Now q/p=(12/9)^k=t
since, 9^k+12^k=16^k Then 1+(4/3)^k=(4/3)^2k Now the equation is t^2-t-1=0 where t=(4/3)^k => t = {1+ sqrt(5)} / 2 and p/q = 9^k / 12^k = (3/4)^k = (4/3)^-k hence 2/(1+(sqrt5)) ans.
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Let,
ReplyDeletelog(9) p = log(12) q = log(16) p+q = k
Hence,
[log p]/2 = k log3 ------1
[log(p+q)]/2 = k log4-------2
log q = k log12--------3
Adding 1 and 2-
[log (p^2 + pq) ]/ 2 = k log12
p^2 + pq = q^2
1+ q/p = (q/p)^2
x^2 - x - 1 = 0
(x-1/2)^2 = 5/4
x = [1+-sqrt(5)]/2
Since x=q/p>0
x= [1+sqrt5]/2
Let the three expressions be equal to k.
ReplyDeleteNow it's easy 9^k+12^k=16^k
We can easily see that 2 satisfies.
So q/p=144/81=16/9
Oops Monday morning blues ..Just ignore my above post:O
ReplyDeleteAh, taking 9^k+12^k=16^k
ReplyDeleteThen 1+(4/3)^k=(4/3)^2k
Now the equation is t^2-t-1=0 where t=(4/3)^k
now t>0 so t=(1+sqrt(5))/2..Now q/p=(12/9)^k=t
since,
ReplyDelete9^k+12^k=16^k
Then 1+(4/3)^k=(4/3)^2k
Now the equation is t^2-t-1=0 where t=(4/3)^k
=> t = {1+ sqrt(5)} / 2
and p/q = 9^k / 12^k = (3/4)^k = (4/3)^-k
hence 2/(1+(sqrt5)) ans.