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YesterdayiSaidTomorow
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Use the properties of logarithms to expand the expression ln z(z1)^2, z>1?
 2 years ago
 2 years ago
YesterdayiSaidTomorow Group Title
Use the properties of logarithms to expand the expression ln z(z1)^2, z>1?
 2 years ago
 2 years ago

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.UserNotFound. Group TitleBest ResponseYou've already chosen the best response.0
Check it on yahoo . It says " = Ln (z) + Ln (z1)^2 = Ln (z) + 2 Ln (z1) "
 2 years ago

YesterdayiSaidTomorow Group TitleBest ResponseYou've already chosen the best response.0
that's the question I posted
 2 years ago

YesterdayiSaidTomorow Group TitleBest ResponseYou've already chosen the best response.0
and no one would explain
 2 years ago

.UserNotFound. Group TitleBest ResponseYou've already chosen the best response.0
amofide xD
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.1
you have rules for logs if you have lnx² that is the same as saying 2lnx you can bring the power to the front of the log
 2 years ago

YesterdayiSaidTomorow Group TitleBest ResponseYou've already chosen the best response.0
but how do you separate it? Like Ln (z) + Ln (z1)^2
 2 years ago

YesterdayiSaidTomorow Group TitleBest ResponseYou've already chosen the best response.0
i mean Ln (z) + 2 Ln (z1)
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.1
also you started with ln z(z1)^2 there is another rule for logs if you have lna + ln b this is equal to lnab so you have ln z(z1)^2 this is being multiplied, so we know we have two logs added together lnz + ln (z1)²
 2 years ago

YesterdayiSaidTomorow Group TitleBest ResponseYou've already chosen the best response.0
Ooooooooooooooh
 2 years ago

YesterdayiSaidTomorow Group TitleBest ResponseYou've already chosen the best response.0
Thanks!
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.1
anytime :D so lna+lnb=lnab lnalnb=lna/b lna^b=blna i think there is another one i forget...
 2 years ago
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