#lnx-4ln3= ln(5/x)#
Recall that #alnb=ln(a^b)#, so
#lnx-ln3^4=ln(5/x)#
Recall that #lna-lnb=ln(a/b)#, so
#ln(x/3^4)= ln(5/x)#
Convert the logarithmic equation to an exponential equation.
#e^ln(x/3^4)= e^ln(5/x)#
Remember that #e^lnx =x#, so
#x/3^4=5/x#
#x^2=5×3^4#
#x=±sqrt(5×3^4) =±sqrt5×sqrt(3^4)=±sqrt5×3^2#
#x=9sqrt5# and #x=-9sqrt5# are possible solutions.
Check:
#lnx-4ln3= ln(5/x)#
If #x=9sqrt5#,
#ln(9sqrt5)-4ln3= ln(5/(9sqrt5))#
#ln9+lnsqrt5-ln3^4 =ln5-ln(9sqrt5)#
#ln9+lnsqrt5-ln(3^2)^2 =ln(sqrt5)^2-ln9-lnsqrt5#
#ln9+lnsqrt5-2ln9=2lnsqrt5-ln9-lnsqrt5#
#lnsqrt5-ln9=lnsqrt5-ln9#
∴ #x=9sqrt5# is a solution.
If #x=-9sqrt5#,
#ln(-9sqrt5)-4ln3= ln(5/(-9sqrt5))=ln5-ln(-9sqrt5)#
But #ln(-9sqrt5)# is not defined.
∴ #x=-9sqrt5# is not a solution.
