• Oct 14th 2008, 03:00 PM
solve ln(x+2)+ln(x-1)=1

it can be written as (x+2)(x-1)=e apparently
What rule is this?

and the answer is x = (sqrt[9+4e]-1)/2
• Oct 14th 2008, 03:07 PM
skeeter
note the sum rule for logs ...

$\log(a) + \log(b) = \log(a \cdot b)$

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

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

change the log equation to an exponential ...

$e^1 = (x+1)(x-1)$
• Oct 14th 2008, 03:12 PM
o ok. how do u change it to an exponential? just put e as the base and it gets rid of ln?

and how did he get that answer? i put it into the quadratic formula but am not getting that same one.
• Oct 14th 2008, 03:13 PM
mr fantastic
Quote:

o ok. how do u change it to an exponential? just put e as the base and it gets rid of ln?

and how did he get that answer? i put it into the quadratic formula but am not getting that same one.

You should know that $\log_A B = C$ is equivalent to $A^C = B$.
• Oct 14th 2008, 03:26 PM
skeeter
$(x+2)(x-1) = e$

$x^2 + x - 2 - e = 0$

$x^2 + x - (2+e) = 0$

$a = 1$, $b = 1$, $c = -(2+e)$

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

$x = \frac{-1 \pm \sqrt{9+4e}}{2}$

since $x > 1$ ...

$x = \frac{-1 + \sqrt{9+4e}}{2}$
• Oct 14th 2008, 03:26 PM
ooooooo very true. thx a lot
• Oct 14th 2008, 03:42 PM
wait, but how do u know x>1?
• Oct 14th 2008, 03:46 PM
skeeter
$\ln(x+2) + \ln(x-1)$ ... what is the domain?
• Oct 14th 2008, 03:48 PM