How do you solve #ln 4 + 2 ln x = 0#?

1 Answer
May 13, 2016

Answer:

I found #x=1/2#

Explanation:

We can first operate on the #2# of the second #ln# to write:
#ln(4)+ln(x)^2=0#
we now use the property of the sum to write:
#ln(4x^2)=0#
we use the definition of log to write:
#4x^2=e^0#
#4x^2=1#
#x^2=1/4#
#x=+-sqrt(1/4)=+-1/2#
where we use only the positive one.