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

Jun 15, 2016

$x = 1 + \sqrt{1 + e}$

Explanation:

Assuming we are looking for Real solutions, we want to find Real values of $x$ such that:

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

Applying the exponential fundtion to both sides, we have:

$e = {e}^{1} = {e}^{\ln \left({x}^{2} - 2 x\right)} = {x}^{2} - 2 x$

Subtracting $e$ from both ends we get:

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

$= \left({x}^{2} - 2 x + 1\right) - \left(e + 1\right)$

$= {\left(x - 1\right)}^{2} - {\left(\sqrt{e + 1}\right)}^{2}$

$= \left(\left(x - 1\right) - \sqrt{e + 1}\right) \left(\left(x - 1\right) + \sqrt{e + 1}\right)$

$= \left(x - \left(1 + \sqrt{e + 1}\right)\right) \left(x - \left(1 - \sqrt{e + 1}\right)\right)$

So $x = 1 \pm \sqrt{e + 1}$

Note that $\sqrt{e + 1} > 1$, so:

• $x = 1 - \sqrt{1 + e} < 0$ making the arguments to $\ln$ in the original equation negative, so having no Real value.

• $x = 1 + \sqrt{1 + e} > 2$ yielding Real values for the natural logarithms in the original equation.

So $x = 1 + \sqrt{1 + e}$ is the only valid Real soltuion of the original equation.

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Footnote

Note that $x = 1 - \sqrt{1 + e}$ is not a solution of the original equation, even if we allow Complex values for $\ln$.

In general:

$\ln \left(z\right) = \ln \left\mid z \right\mid + A r g \left(z\right) i$

So in the case of Real negative values of $z$ we find:

$\ln \left(z\right) = \ln \left\mid z \right\mid + \pi i$

In particular:

$\ln \left(1 - \sqrt{1 + e}\right) + \ln \left(1 - \sqrt{1 + e} - 2\right)$

$= \ln \left(\sqrt{1 + e} - 1\right) + \pi i + \ln \left(\sqrt{1 + e} + 1\right) + \pi i$

$= \ln \left(\left(\sqrt{1 + e} - 1\right) \left(\sqrt{1 + e} + 1\right)\right) + 2 \pi i$

$= \ln \left(e\right) + 2 \pi i$

$= 1 + 2 \pi i \ne 1$