SOLUTION: {{{ (log((z)))^2=log((z^2)) }}}

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: {{{ (log((z)))^2=log((z^2)) }}}      Log On


   

Question 630604: ++%28log%28%28z%29%29%29%5E2=log%28%28z%5E2%29%29++
Found 2 solutions by lwsshak3, stanbon:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
(log(z))^2=log(z^2)
(log(z))^2-log(z^2)=0
(log(z))^2-2log(z)=0
log(z)(log(z)-2)=0
log(z)=0
z=1
or
log(z)-2=0
log(z)=2
z=100

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(log((z)))^2=log((z^2))
-----
= [log(z)]^2 - 2log(z) = 0
----
Factor:
log(z)[log(z)-2)] = 0
----
log(z) = 0 or log(z) = 2
----
z = 1 or z = 100
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Cheers,
Stan H.
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