SOLUTION: hello. i need to simplify ln(z^4-z^2). i thought that i would use division because of the log rules but i keep getting the incorrect answer. ln(z^4)/ln(z^2) apparently isnt the ans

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: hello. i need to simplify ln(z^4-z^2). i thought that i would use division because of the log rules but i keep getting the incorrect answer. ln(z^4)/ln(z^2) apparently isnt the ans      Log On


   

Question 651799: hello. i need to simplify ln(z^4-z^2). i thought that i would use division because of the log rules but i keep getting the incorrect answer. ln(z^4)/ln(z^2) apparently isnt the answer and i do not know what is...
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
ln(z^4-z^2)
-----
Factor:
= ln(z^2(z^2-1))
-----
= ln[z^2(z-1)(z+1)]
----
= ln(z^2) + ln(z-1) + ln(z+1)
-----
= 2ln(z) + ln(z-1) + ln(z+1)
===============================
Cheers,
Stan H.
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Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
ln(z^4-z^2)

ln(z^2(z^2-1))

ln(z^2(z-1)(z+1))

ln(z^2) + ln(z-1) + ln(z+1)

2ln(z) + ln(z-1) + ln(z+1)

So the answer is

2ln(z) + ln(z-1) + ln(z+1)