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ksaimouli
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dw:1361407430942:dw

ksaimouli
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how this gets to \[\ln \sqrt{2}\]

ksaimouli
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@TuringTest

tkhunny
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ln(1) = 0

ksaimouli
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i know that so dw:1361408007302:dw

tkhunny
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You MUST have been given fundamental principles of logarithms,
Here's one: log(a) + log(b) = log(a*b)
Have you any others?

ksaimouli
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i dont get this

ksaimouli
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this is  so division right

tkhunny
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Good. log(a)  log(b) = log(a/b)
Any more?

ksaimouli
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dw:1361408413459:dw

ksaimouli
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so ln(2)/(2)

ksaimouli
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how did they got ln sqrt 2

ksaimouli
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@SithsAndGiggles

tkhunny
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So, you're saying you have only two properties for logarithms? You don't have this one: \(a\cdot \log(b) = \log\left(b^{a}\right)\)?

ksaimouli
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dw:1361408835272:dw where do u use that

tkhunny
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\(\dfrac{\ln(2)}{2} = \dfrac{1}{2}\cdot \ln(2) = ln\left(2^{1/2}\right) = \ln(\sqrt{2})\)
It's all in the properties!

ksaimouli
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wow thx