help understand answer

**Shadowfox1** · October 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

**skeeter** · October 14th 2008, 03:07 PM

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 ...

$e^1 = (x+1)(x-1)$

**Shadowfox1** · October 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.

**mr fantastic** · October 14th 2008, 03:13 PM

Originally Posted by Shadowfox1

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 $A^C = B$ .

**skeeter** · October 14th 2008, 03:26 PM

$(x+2)(x-1) = e$

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

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

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

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

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

since $x > 1$ ...

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

**Shadowfox1** · October 14th 2008, 03:26 PM

ooooooo very true. thx a lot

**Shadowfox1** · October 14th 2008, 03:42 PM

wait, but how do u know x>1?

**skeeter** · October 14th 2008, 03:46 PM

$\ln(x+2) + \ln(x-1)$ ... what is the domain?

**Shadowfox1** · October 14th 2008, 03:48 PM

o ok, i see. I can never find the domain without a calculator.

**skeeter** · October 14th 2008, 03:57 PM

I can never find the domain without a calculator.

learn the concept of domain of a function ... the "machine" will put you at a great disadvantage.

**Shadowfox1** · October 14th 2008, 03:58 PM

yea true, i should. its just hard when you get stuff involving e and log and stuff like that

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