Question 110930
Let {{{w=ln(x)}}}


So we get {{{w=ln(x^9)}}}


{{{w^3=9*ln(x)}}} Rewrite the logarithm using the identity {{{ln(x^a)=a*ln(x)}}}



{{{w^3=9*w}}} Replace {{{ln(x)}}} with w



{{{w^3-9*w=0}}} Subtract 9w from both sides


{{{w(w^2-9)=0}}} Factor out a w




{{{w(w+3)(w-3)=0}}} Factor using the difference of squares



So our solutions are (in terms of w)


{{{w=0}}}, {{{w=-3}}}, or {{{w=3}}}



So this means {{{ln(x)=0}}}, {{{ln(x)=-3}}}, or {{{ln(x)=3}}}



Now raise all sides as an exponent with base e



{{{e^(ln(x))=e^0}}}, {{{e^(ln(x))=e^(-3)}}}, or  {{{e^(ln(x))=e^3}}}


Simplify


{{{x=1}}}, {{{x=1/e^3}}}, or  {{{x=e^3}}}