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

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Answer 1 · 2016-07-21T16:02:27

$x = \frac{1 + \sqrt{1 + 4 e}}{2} = 2.22287$ $x = (1 + sqrt (1 + 4 e ))/2=2.22287$ , nearly.

$x > 1$ $x > 1$ to make both $ln x and ln (x - 1)$ $ln x and ln (x - 1 )$ real.

ln x + ln ( x - 1 ) = ln (x(x-1)) = 1 = ln e#.

So, $x (x - 1) = e$ $x(x-1) = e$ . Solving,

$x . = . \frac{1 \pm \sqrt{1 + 4 e}}{2}$ $x .=.(1+-sqrt(1+4e))/2$

Negative root is inadmissible.

The positive root is 2,22287, nearly .

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