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

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Answer 1 · 2016-04-27T00:52:17

Use the log rule #log_an - log_am = log_a(n / m)#

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

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

#x^2 + 4x = 3(x^3 - x)#

#x^2 + 4x = 3x^3 - 3x#

#0 = 3x^3 + x^2 - 7x#

#0 = x(3x^2 + x - 7)#

Solving the inner trinomial by the quadratic formula:

#x = (-1 +- sqrt(1^2 - (4 xx 3 xx -7)))/(2 xx 3)#

#x = (-1 +- sqrt(85))/6 and 0#

Hopefully this helps!