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

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- 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 ...

change the log equation to an exponential ...

- 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

, ,

since ...

- 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
... 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