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

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Answer 1 · 2015-11-14T02:34:26

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

Using the rules of logarithms,

$ln(x)+ln(x-1)=ln(x*(x-1))=ln(x^2-x)$ .

Therefore,

$ln(x^2-x)=1$ .

Then, we exponentiate both sides (put both sides to the $e$ power):

$e^(ln(x^2-x))=e^1$ .

Simplify, remembering that exponents undo logarithms:

$x^2-x=e$ .

Now, we complete the square:

$x^2-x+1/4=e+1/4$

Simplify:

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

Take the square root of both sides:

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

Add $1/2$ to both sides:

$x=(1±sqrt(4e+1))/2$

Eliminate the negative answer (in $log_"a"b, b>0$ ):

$=> color(blue)(x=(1+sqrt(4e+1))/2)$

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