ln(x) + ln(x-1) = 1

Solve for x

Printable View

- Sep 9th 2008, 03:58 PMdm10Solve for x
ln(x) + ln(x-1) = 1

Solve for x - Sep 9th 2008, 04:28 PMmr fantastic
- Sep 9th 2008, 04:32 PMProve It
First off, we can't have $\displaystyle x\leq 0$

Remember the rule

$\displaystyle \log_a(m)+\log_a(n)=\log_a(mn)$?

Using this rule we get

$\displaystyle \ln[x(x-1)]=1$

$\displaystyle x(x-1)=e^1$

$\displaystyle x(x-1)=e$

$\displaystyle x^2-x=e$

$\displaystyle x^2-x-e=0$

Then using the quadratic formula, namely

If $\displaystyle ax^2 + bx + c = 0$ then $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

we get as our solution for x...

$\displaystyle x=\frac{1\pm \sqrt{(-1)^2-4\times 1 \times (-e)}}{2\times 1}$

$\displaystyle =\frac{1 \pm \sqrt{1+4e}}{2}$

And since we know $\displaystyle x > 0$ we disregard the negative answer.

So $\displaystyle x = \frac{1 + \sqrt{1 + 4e}}{2}$. - Sep 9th 2008, 04:33 PMdm10
- Sep 9th 2008, 04:34 PMProve It