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How do you solve # log x^2 = (log x)^2#?

1 Answer

Vinícius Ferraz

Dec 1, 2015

#1 or e^2#

Explanation:

#y = log x#

#2y = y^2#

#y = 0 or y = 2#

#log x = 0 or log x = 2#

#x = e^0 or x = e^2#

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