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Mathematics 20 Online
OpenStudy (anonymous):

Can you show me why logn^2 = 2log2n?

OpenStudy (anonymous):

\[\log_{}n ^{2} = 2\log_{2}n \]

OpenStudy (dumbcow):

its an identity \[\log (n^{a}) = a \log(n)\] you could also prove it using product identity \[\log(n^{2}) = \log(n*n) = \log(n) +\log(n) = 2\log(n)\]

OpenStudy (anonymous):

ah ok, but why does the base change from base 10 to base 2?

OpenStudy (anonymous):

or I think maybe the 1st one is actually base 2 as well...

OpenStudy (dumbcow):

umm yeah identities only work with like bases...

OpenStudy (amistre64):

log(n^2) log(n*n) log(n) + log(n) 2 log(n)

OpenStudy (amistre64):

thrms are not accepted on faith ... they are proven or rejected

OpenStudy (anonymous):

ok thanks everyone

OpenStudy (anonymous):

thanks too amistre

OpenStudy (amistre64):

yep :)

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