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

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

YesterdayiSaidTomorow
 3 years ago
Best ResponseYou've already chosen the best response.0that's the question I posted

YesterdayiSaidTomorow
 3 years ago
Best ResponseYou've already chosen the best response.0and no one would explain

amorfide
 3 years ago
Best ResponseYou've already chosen the best response.1you 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

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

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

amorfide
 3 years ago
Best ResponseYou've already chosen the best response.1also 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

YesterdayiSaidTomorow
 3 years ago
Best ResponseYou've already chosen the best response.0Ooooooooooooooh

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