How do you solve #log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1#?

1 Answer
Jun 11, 2016

Answer:

#x = 1/6 (1 + sqrt[85])#

Explanation:

#log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = -1# or
#log_(1/3) (x^2 + 4x) - log_(1/3) (x^3 - x) = log_(1/3)(1/3)^(-1)#
then

# (x^2 + 4x)/ (x^3 - x) =3# or # (x + 4)/ (x^2 - 1) =3#

or

#3x^2-3=x+4# or finally #3x^2-x-7=0#

solving for #x# we get #{x = 1/6 (1 - sqrt[85])}, {x = 1/6 (1 + sqrt[85])}# but we choose #x = 1/6 (1 + sqrt[85])# because #x = 1/6 (1 - sqrt[85]) <0# and makes #(x^3 - x) < 0#