How do you solve #(2lnx) + 1 = ln(2x)#?

1 Answer
sente
Nov 30, 2015

# x = 2/e#

Explanation:

We will use the following:

  • #ln(a^x) = xln(a)#
  • #ln(a) - ln(b) = ln(a/b)#
  • #e^ln(a) = a#

#2ln(x) + 1 = ln(2x)#

#=> ln(x^2) + 1 = ln(2x)# (by the first property above)

#=> ln(x^2) - ln(2x) = -1#

#=> ln(x^2/(2x)) = -1# (by the second property above)

#=> ln(x/2) = -1#

#=> e^ln(x/2) = e^(-1)#

#=> x/2 = 1/e# (by the third property above)

#:. x = 2/e#

Related questions
See all questions in Logarithmic Models
Impact of this question
9960 views around the world
You can reuse this answer
Creative Commons License