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

1 Answer
Dec 2, 2015

Answer:

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

Explanation:

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#