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

How do you solve $\mathrm{ln}x+\mathrm{ln}\left(x-1\right)=1$ ?
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MoxboasteBots5h
Using the rules of logarithms,
$\mathrm{ln}\left(x\right)+\mathrm{ln}\left(x-1\right)=\mathrm{ln}\left(x\cdot \left(x-1\right)\right)=\mathrm{ln}\left({x}^{2}-x\right)$
Therefore,
$\mathrm{ln}\left({x}^{2}-x\right)=1$
Then, we exponentiate both sides (put both sides to the e power):
${e}^{\mathrm{ln}\left({x}^{2}-x\right)}={e}^{1}$
Simplify, remembering that exponents undo logarithms:
${x}^{2}-x=e$
Now, we complete the square:
${x}^{2}-x+\frac{1}{4}=e+\frac{1}{4}$
Simplify:
${\left(x-\frac{1}{2}\right)}^{2}=e+\frac{1}{4}=\frac{4e+1}{4}$
Take the square root of both sides:
$x-\frac{1}{2}=\frac{±\sqrt{4e+1}}{2}$
Add $\frac{1}{2}$ to both sides:
$x=\frac{1±\sqrt{4e+1}}{2}$
Eliminate the negative answer (in ${\mathrm{log}}_{a}b,b>0$):
$⇒x=\frac{1+\sqrt{4e+1}}{2}$

Joseph Lewis
Simplify the left side.
$\mathrm{ln}\left({x}^{2}-x\right)=1$
To solve for x, rewrite the equation using properties of logarithms.
${e}^{\mathrm{ln}\left({x}^{2}-x\right)}={e}^{1}$
Solve for x
$x=\frac{1+\sqrt{1+4e}}{2},\frac{1-\sqrt{1+4e}}{2}$
Exclude the solutions that do not make $\mathrm{ln}\left(x\right)+\mathrm{ln}\left(x-1\right)=1$ true.
$x=\frac{1+\sqrt{1+4e}}{2}$
The result can be shown in multiple forms.
Exact Form:
$x=\frac{1+\sqrt{1+4e}}{2}$
Decimal Form:
$x=2.22287-22\dots$