Question

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

Answer 1

`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(x) + ln(x-2) = ln(x(x-2)) = ln(x^2-2x)`

Applying the exponential fundtion to both sides, we have:

`e = e^1 = e^(ln(x^2-2x)) = x^2-2x`

Subtracting `e` from both ends we get:

`0 = x^2-2x-e`

`= (x^2-2x+1)-(e+1)`

`= (x-1)^2-(sqrt(e+1))^2`

`= ((x-1)-sqrt(e+1))((x-1)+sqrt(e+1))`

`= (x-(1+sqrt(e+1)))(x-(1-sqrt(e+1)))`

So `x=1+-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(z) = ln abs(z) + Arg(z)i`

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

`ln(z) = ln abs(z) + pii`

In particular:

`ln(1-sqrt(1+e))+ln(1-sqrt(1+e)-2)`

`=ln(sqrt(1+e)-1) + pii + ln(sqrt(1+e)+1) + pii`

`=ln((sqrt(1+e)-1)(sqrt(1+e)+1)) + 2pii`

`=ln(e) + 2pii`

`=1 + 2pii != 1`