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Old 04-30-2008, 09:25 AM
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Default logarithms prob..simple but blur

help me on this problem..tq for your time
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Old 04-30-2008, 09:27 AM
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Hello,

Your last step is ok !

Now \log \frac 1a=-\log(a)

Does that help ?
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Old 04-30-2008, 09:40 AM
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Quote:
Originally Posted by Moo View Post
Hello,

Your last step is ok !

Now \log \frac 1a=-\log(a)

Does that help ?
yap tq for your kindness Moo...
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