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secondary 4 | A Maths
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Date Posted: 3 years ago
Views: 185
log2(3) x log3(4) x ... x logp(p+1) = 2logp(p⁴) + logp(1)
logp(3)/logp(2) x logp(4)/logp(3) x... x logp(p-1)/logp(p-2) x logp(p)/logp(p-1) x logp(p+1)/logp(p) = 2 x 4 logp(p) + 0
Common terms cancel out. We are left with :
logp(p+1)/logp(2) = 8
log2(p+1) = 8
p+1 = 2^8 = 256
p = 255
logp(3)/logp(2) x logp(4)/logp(3) x... x logp(p-1)/logp(p-2) x logp(p)/logp(p-1) x logp(p+1)/logp(p) = 2 x 4 logp(p) + 0
Common terms cancel out. We are left with :
logp(p+1)/logp(2) = 8
log2(p+1) = 8
p+1 = 2^8 = 256
p = 255
Amazing! U r a maths genius
No, this is not difficult at all. Why you delete your amended answer by the way? It was correct
I thought, I am wrong
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We change the bases to a common base for the products on the left hand side so we can do lots of cancellations.
Date Posted:
3 years ago
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