How do you solve #ln x - 4 ln 3 = ln (5 / x)#?

1 Answer
Aug 13, 2015

#color(red)(x=9sqrt5)#

Explanation:

#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.