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

Jul 15, 2016

$x = \left\{- 1 , - \frac{1}{3}\right\}$

Explanation:

${\log}_{x + 3} \left(\frac{{x}^{3} + x - 2}{x}\right) = \frac{{\log}_{e} \left(\frac{{x}^{3} + x - 2}{x}\right)}{\log} _ e \left(x + 3\right) = 2$

so

${\log}_{e} \left(\frac{{x}^{3} + x - 2}{x}\right) = {\log}_{e} {\left(x + 3\right)}^{2}$

or

$\left({x}^{3} + x - 2\right) = x {\left(x + 3\right)}^{2}$

simplifying

$3 {x}^{2} + 4 x + 1 = 0$

obtaining for $x$ the values

$\left\{- 1 , - \frac{1}{3}\right\}$

and both are feasible solutions.

Jul 15, 2016

$x = - \frac{1}{3} \mathmr{and} x = - 1$

Explanation:

By definition, if ${\log}_{a} b = c \text{, then } {a}^{c} = b$

So in this case

if ${\log}_{x + 3} \left(\frac{{x}^{3} + x - 2}{x}\right) = 2 , \text{then } {\left(x + 3\right)}^{2} = \frac{{x}^{3} + x - 2}{x}$

${\left(x + 3\right)}^{2} = \frac{{x}^{3} + x - 2}{x}$

${x}^{2} + 6 x + 9 = \frac{{x}^{3} + x - 2}{x} \text{ now } \times x$

${x}^{3} + 6 {x}^{2} + 9 x = {x}^{3} + x - 2 \text{ cancel " x^3 " on each side}$

$6 {x}^{2} + 9 x = x - 2$

$6 {x}^{2} + 8 x + 2 = 0 \text{ } \div 2$

=$3 {x}^{2} + 4 x + 1 = 0$

$\left(3 x + 1\right) \left(x + 1\right) = 0$

$3 x + 1 = 0 \Rightarrow x = - \frac{1}{3} \text{ or } x + 1 = 0 \Rightarrow x = - 1$

$x = - \frac{1}{3} \mathmr{and} x = - 1$