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

1 Answer
Oct 5, 2015

#x = +-10#

Explanation:

Assuming that's the common log, take the base 10 exponential of both sides, that is, the "sideth" power of 10.

#log(x^2) = 2#
#x^2 = 10^2#

Take the root

#x = +-10#

Since #x^2# is positive for all values of #x# the only value of #x# we can't have is #0#, so both answers are okay.