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

Dec 12, 2015

Use the rule which states that the sum of two logarithms is the logarithm of the product, and that the difference of two logarithm is the logarithm of the ratio.

Explanation:

For the first rule, you have that

$\ln \left(x - 1\right) + \ln \left(x\right) = \ln \left(x \left(x - 1\right)\right)$

For the second rule, you have that

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

This expression can be further simplified, since the solutions of

${x}^{2} + x - 2 = 0$

are $1$ and $- 2$, we have that

${x}^{2} + x - 2 = \left(x - 1\right) \left(x + 2\right)$, so we have

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

And so the final answer is $\ln \left(\setminus \frac{x}{x + 2}\right)$