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

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Answer 1 · 2016-07-21T16:02:27

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

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

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

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

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

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?