# How do you solve ln ( x - 1) + ln x - ln ( x^2 + x - 2 ) = 0?

Dec 17, 2015

The equation has no solution.

#### Explanation:

Know the following rule: $\ln \left(a\right) + \ln \left(b\right) - \ln \left(c\right) = \ln \left(\frac{a b}{c}\right)$

Basically, if logarithms with the same base are being added, the terms on the interior can be multiplied. If they're being subtracted, the terms are divided.

Thus, the equation can be simplified as

$\ln \left(\frac{x \left(x - 1\right)}{{x}^{2} + x - 2}\right) = 0$

Factor the denominator.

$\ln \left(\frac{x \left(x - 1\right)}{\left(x + 2\right) \left(x - 1\right)}\right) = 0$

Simplify.

$\ln \left(\frac{x}{x + 2}\right) = 0$

Rewrite, recalling that $\ln \left(a\right) = {\log}_{e} \left(a\right)$.

${e}^{0} = \frac{x}{x + 2}$

$1 = \frac{x}{x + 2}$

$x + 2 = x$

$0 = 2$

This equation has no solutions.

This can be confirmed by checking a graph.

graph{ln(x-1)+ln(x)-ln(x^2+x-2) [-1.29, 18.71, -3.8, 6.2]}

The graph approaches the $x$-axis but is never actually equal to $0$.