How do you solve # (ln x)^2=ln x^2#?

1 Answer
Jun 23, 2016

Answer:

#x=1color(white)("XXX")orcolor(white)("XXX")x=e^2#

Explanation:

Remember: #ln(x^2)=2ln(x)#

Let #k=ln(x)#

Therefore
#color(white)("XXX")(ln(x))^2=ln(x^2)#
is equivalent to
#color(white)("XXX")k^2=2k#

#color(white)("XXX")k^2-2k=0#

#color(white)("XXX")k(k-2)=0#

#color(white)("XXX"){: (k=0,color(white)("XX")orcolor(white)("XX"),k=2), (rarr ln(x)=0,,rarr ln(x)=2), (rarr e^0=x,,rarr e^2=x), (rarr x=1,,rarr x=e^2) :}#