How do you solve #(\ln ( x ) ) ^ { 2} + 2\ln ( x ) = + 15#?

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Answer 1 · 2016-11-17T16:00:26

#x=e^((-5))color(white)(aaa)x=e^3#
or
#x=.0067379color(white)(aaa)x=20.0855#

#(ln(x))^2 +2ln(x)=15#

Subtract 15 from both sides.

#(lnx)^2+2lnx-15=0#

Factor (just factor as if the equation were #x^2+2x-15=0#, then use #lnx# in place of #x#).

#(lnx+5)(lnx-3)=0#

Set each factor equal to zero and solve.

#lnx+5=0color(white)(aaa)lnx-3=0#

#lnx=-5color(white)(aaa)lnx=3#

Rewrite as an exponential. The base of #ln# is #e#.

#x=e^((-5))color(white)(aaa)x=e^3#

#x=.0067379color(white)(aaa)x=20.0855#