Question

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

Answer 1

`x= (1+sqrt(1+4e))/2 = "(approx.)" 2.22287`

Explanation:

If
`color(white)("XXX")ln(x)+ln(x-1)=1`
then
`color(white)("XXX")e^(ln(x)+ln(x-1))= e^1`

`rarrcolor(white)("XXX")e^(ln(x))*e^(ln(x-1)) = e^1`

`rarrcolor(white)("XXX")x*(x-1) = e`

`rarrcolor(white)("XXX")x^2-x-e=0`

Using the quadratic formula
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`

`x= (1+-sqrt(1+4e))/2`

Since `(sqrt(1+4e) > 1)`
`color(white)("XXXXXXXXXXX")(1-sqrt(1+4e))/2` is negative
and
Since `ln(x)` is undefined for `x < 0`
`color(white)("XXX")x=(1-sqrt(1+4e))/2` is an extraneous solution
and only
`color(white)("XXX")x=(1+sqrt(1+4e))/2` is valid