# Solve for x

• Sep 9th 2008, 03:58 PM
dm10
Solve for x
ln(x) + ln(x-1) = 1

Solve for x
• Sep 9th 2008, 04:28 PM
mr fantastic
Quote:

Originally Posted by dm10
ln(x) + ln(x-1) = 1

Solve for x

Use the following rules:

1. $\displaystyle \ln A + \ln B = \ln (AB)$.

2. $\displaystyle \ln C = D \Rightarrow e^D = C$

After applying the above rules, expand the result and arrange into a quadratic equation which you can solve for x.
• Sep 9th 2008, 04:32 PM
Prove It
Quote:

Originally Posted by dm10
ln(x) + ln(x-1) = 1

Solve for x

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 PM
dm10
Quote:

Originally Posted by mr fantastic
Use the following rules:

1. $\displaystyle \ln A + \ln B = \ln (AB)$.

2. $\displaystyle \ln C = D \Rightarrow e^D = C$

After applying the above rules, expand the result and arrange into a quadratic equation which you can solve for x.

Thanks for the help.
• Sep 9th 2008, 04:34 PM
Prove It
Quote:

Originally Posted by dm10
I did use the rules but I came out with the wrong answer. The answer I got was -1/2 but the correct answer is 1/2(1 + (the square root of) 1 + 4e)

See my response above. Can you see where you've gone wrong?