Question

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

Answer 1

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

Explanation:

`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

`x = -1/3 or x= -1`

Explanation:

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`