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

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

YesterdayiSaidTomorowBest ResponseYou've already chosen the best response.0
that's the question I posted
 one year ago

YesterdayiSaidTomorowBest ResponseYou've already chosen the best response.0
and no one would explain
 one year ago

amorfideBest 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
 one year ago

YesterdayiSaidTomorowBest ResponseYou've already chosen the best response.0
but how do you separate it? Like Ln (z) + Ln (z1)^2
 one year ago

YesterdayiSaidTomorowBest ResponseYou've already chosen the best response.0
i mean Ln (z) + 2 Ln (z1)
 one year ago

amorfideBest 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
 one year ago

YesterdayiSaidTomorowBest ResponseYou've already chosen the best response.0
Ooooooooooooooh
 one year ago

amorfideBest 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...
 one year ago
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