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 Post subject: Log Problem Help!
PostPosted: Tue, 27 May 2003 23:20:12 UTC 
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Joined: Tue, 27 May 2003 23:15:18 UTC
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Please show all steps thank you!

If log (base"2n") 1944 = log (base"n") 486(2)^(1/2)

determine the value of n^6.


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 Post subject:
PostPosted: Wed, 28 May 2003 01:28:42 UTC 
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Joined: Mon, 19 May 2003 07:15:29 UTC
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albertgates wrote:
Please show all steps thank you!

If log (base"2n") 1944 = log (base"n") 486(2)^(1/2)

determine the value of n^6.


Now sure what you mean by the 486(2)^(1/2) there

Did you mean
\log_{2n}1944=\sqrt{2}\log_{n}486

\log_{2n}1944=\frac{\log_{n}1944}{\log_{n}(2n)}=\sqrt{2}\log_{n}486
\frac{\log_{n}1944}{\log_{n}486}=\sqrt{2}\log_{n}(2n)
\frac{\log_{n}1944}{\sqrt{2}\log_{n}486}=\log_{n}n+\log_{n}2
\frac{6\log_{n}1944}{\sqrt{2}\log_{n}486}=\log_{n}(n^6)+\log_{n}64
Let
p=\frac{6\log_{n}1944}{\sqrt{2}\log_{n}486}-\log_{n}64
Then
n^6=n^p

Of course, since I didn't understand your question, then my result is probably not what you wanted.

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