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

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Answer 1 · 2015-12-02T07:47:29

Apply the exponential function and use a property of logarithms to find that
#x = +-e#

For this problem, we will use that
#e^log(a) = a#

#log(x^2) = 2#

#=> e^log(x^2) = e^2#

#=> x^2 = e^2#

#=> sqrt(x^2) = sqrt(e^2)#

#=> |x| = e#

#=> x = +-e#