Here's the question you clicked on:
YesterdayiSaidTomorow
Use the properties of logarithms to expand the expression ln z(z-1)^2, z>1?
Check it on yahoo . It says " = Ln (z) + Ln (z-1)^2 = Ln (z) + 2 Ln (z-1) "
that's the question I posted
and no one would explain
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
but how do you separate it? Like Ln (z) + Ln (z-1)^2
i mean Ln (z) + 2 Ln (z-1)
also you started with ln z(z-1)^2 there is another rule for logs if you have lna + ln b this is equal to lnab so you have ln z(z-1)^2 this is being multiplied, so we know we have two logs added together lnz + ln (z-1)²
Ooooooooooooooh
anytime :D so lna+lnb=lnab lna-lnb=lna/b lna^b=blna i think there is another one i forget...