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

Printable View

- Oct 14th 2008, 03:00 PMShadowfox1help understand answer
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 PMskeeter
note the sum rule for logs ...

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

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

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

change the log equation to an exponential ...

$\displaystyle e^1 = (x+1)(x-1)$ - Oct 14th 2008, 03:12 PMShadowfox1
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 PMmr fantastic
- Oct 14th 2008, 03:26 PMskeeter
$\displaystyle (x+2)(x-1) = e$

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

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

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

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

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

since $\displaystyle x > 1$ ...

$\displaystyle x = \frac{-1 + \sqrt{9+4e}}{2}$ - Oct 14th 2008, 03:26 PMShadowfox1
ooooooo very true. thx a lot

- Oct 14th 2008, 03:42 PMShadowfox1
wait, but how do u know x>1?

- Oct 14th 2008, 03:46 PMskeeter
$\displaystyle \ln(x+2) + \ln(x-1)$ ... what is the domain?

- Oct 14th 2008, 03:48 PMShadowfox1
o ok, i see. I can never find the domain without a calculator.

- Oct 14th 2008, 03:57 PMskeeterQuote:

I can never find the domain without a calculator.

- Oct 14th 2008, 03:58 PMShadowfox1
yea true, i should. its just hard when you get stuff involving e and log and stuff like that