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

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Answer 1 · 2016-07-15T20:25:50

#x = {-1,-1/3}#

#log_[x+3] ((x^3+x-2)/(x))=(log_e ((x^3+x-2)/x))/log_e(x+3) =2#

so

#log_e ((x^3+x-2)/x) = log_e(x+3)^2#

or

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

simplifying

#3x^2+4x+1=0#

obtaining for #x# the values

#{-1,-1/3}#

and both are feasible solutions.

Answer 2 · 2016-07-15T21:15:08

#x = -1/3 or x= -1#

By definition, if #log_a b = c", then "a^c = b#

So in this case

if #log_(x+3) ((x^3 +x -2)/x) = 2, "then " (x+3)^2 = (x^3 +x -2)/x#

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

#x^2 +6x +9 = (x^3 +x -2)/x " now "xx x#

#x^3 +6x^2 +9x = x^3 +x -2 " cancel " x^3 " on each side"#

#6x^2 +9x = x -2#

#6x^2 +8x +2 = 0 " " div2#

=#3x^2 +4x +1 = 0#

#(3x +1)(x+1) = 0#

#3x +1 =0 rArr x = -1/3" or " x + 1 = 0 rArr x =-1#

#x = -1/3 or x= -1#