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

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Answer 1 · 2015-11-14T02:34:26

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

Using the rules of logarithms,

$ln (x) + ln (x - 1) = ln (x \cdot (x - 1)) = ln (x^{2} - x)$ $ln(x)+ln(x-1)=ln(x*(x-1))=ln(x^2-x)$ .

Therefore,

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

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

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

Simplify, remembering that exponents undo logarithms:

$x^{2} - x = e$ $x^2-x=e$ .

Now, we complete the square:

$x^{2} - x + \frac{1}{4} = e + \frac{1}{4}$ $x^2-x+1/4=e+1/4$

Simplify:

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

Take the square root of both sides:

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

Add $\frac{1}{2}$ $1/2$ to both sides:

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

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

$\Rightarrow x = \frac{1 + \sqrt{4 e + 1}}{2}$ $=> color(blue)(x=(1+sqrt(4e+1))/2)$

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