Solve $\ln(x)+\ln(x-1)=0$ for $x$

Question

Solve the following equation for x;

$$\ln(x)+\ln(x-1)=0$$

What I did is the following but I'm pretty sure its wrong..

$$\ln(x)+\ln(x-1)=0$$ $$\ln(x)=-\ln(x-1)$$ $$e^{\ln(x)}=e^{-\ln(x-1))}$$ $$x=-x-1$$ $$2x=-1$$ $$x=-\frac{1}{2}$$

Answer 1

Hint

$$\ln a+\ln b=\ln(ab)$$ and

$$\ln a=0\iff a=1$$

Answer 2

one of the correct ways is:

$$ \ln(x)+\ln(x-1)=0 \iff \ln(x)=-\ln(x-1)\overset{note\ *^1}{\iff} \ln(x)=\ln\Big({1 \over x-1}\Big)\\ \ln(x)=\ln\Big({1 \over x-1}\Big) \iff \begin{cases} x={1 \over x-1}\\x-1>0\\x>0 \end{cases} \iff \begin{cases} x^2-x-1=0\\x>1 \end{cases} \iff x=\frac{\sqrt{5}+1}{2} $$

ALTERNATIVE: (shorter)

$$ \ln(x)+\ln(x-1)=0 \iff \begin{cases} \ln(x(x-1))=0\\x-1>0\\x>0 \end{cases} \iff \\ \iff \begin{cases} x(x-1)=1\\x>1 \end{cases} \iff x=\frac{\sqrt{5}+1}{2} $$

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all in all the most remarkable thing about that equation is that it has the golden ratio as his solution.

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note $*^1$:
$ -\ln(a)=\ln(1)-\ln(a)=\ln\Big({1 \over a}\Big) $

Answer 3

The obvious way:

$0 =\ln (x) + \ln(x-1) =\ln(x(x-1)) $.

Taking $\exp$ of both sides, $1 = x(x-1)$ or $x^2-x-1=0$ so $x =\dfrac{1\pm\sqrt{1+4}}{2} =\dfrac{1\pm\sqrt{5}}{2} $

Answer 4

You've made a small error in your post. It's this part here:

$$ -\ln(x-1)=\ln[(x-1)^{-1}]=\ln{\frac 1 {x-1}} $$

Because $a \ln b = \ln{b^a}$. So,

$$ e^{-\ln(x-1)}=\frac 1 {x-1} $$

Having established that and using your exact method of solution it would be:

$$ \ln(x)+\ln(x-1)=0 \\ \ln(x) = -\ln(x-1) \\ e^{\ln(x)} = e^{-\ln(x-1)} \\ x = \frac 1 {x-1} \\ (x-1)x=1 \\ x^2-x-1=0 \\ x = \frac {1 \pm \sqrt 5}{2} $$

Of course, $x>1$ for the logarithms to be defined so one solution does not satisfy the initial equation. So,

$$ x = \frac {1 + \sqrt 5}{2} $$