# ln and e

• Jan 14th 2012, 05:25 PM
Furyan
ln and e
The question is:

Given that $\ln(2x^2+9x-5)=1+\ln(x^2+2x-15)$, find $x$ in terms of $e$

I got as far as:

$e = \dfrac{2x-1}{x-3}$

The next step is

$e = \dfrac{2x-1}{x-3} \Rightarrow 3e - 1 = x(e - 2)$

I'm probably just being dense, but I don't know how to do that step. Could somebody please help me.

Thank you
• Jan 14th 2012, 05:56 PM
icemanfan
Re: ln and e
$x = \frac{3e - 1}{e - 2}$

• Jan 14th 2012, 06:05 PM
Furyan
Re: ln and e
Quote:

Originally Posted by icemanfan
$x = \frac{3e - 1}{e - 2}$

Thanks for your reply. I see the last step, but I don't know how to do the step I've shown in original post.
• Jan 14th 2012, 10:59 PM
Quacky
Re: ln and e
Quote:

Originally Posted by Furyan
The question is:

Given that $\ln(2x^2+9x-5)=1+\ln(x^2+2x-15)$, find $x$ in terms of $e$

I got as far as:

$e = \dfrac{2x-1}{x-3}$

The next step is

$e = \dfrac{2x-1}{x-3} \Rightarrow 3e - 1 = x(e - 2)$

I'm probably just being dense, but I don't know how to do that step. Could somebody please help me.

Thank you

$e = \dfrac{2x-1}{x-3}$

Multiply through by $(x-3)$:

$e(x-3)=2x-1$

Expand:

$ex-3e=2x-1$

Rearrange, so that terms involving an " $x$" are isolated on one side of the equation:

$ex-2x=3e-1$

Factor:

$x(e-2)=3e-1$
• Jan 15th 2012, 04:05 AM
Furyan
Re: ln and e
Quote:

Originally Posted by Quacky
$e = \dfrac{2x-1}{x-3}$

Multiply through by $(x-3)$:

$e(x-3)=2x-1$

$ex-3e=2x-1$

Rearrange, so that terms involving an " $x$" are isolated on one side of the equation:

$ex-2x=3e-1$

Factor:

$x(e-2)=3e-1$

Thanks Quacky for taking the time to reply, that was what I was asking. Normally I'd see that, I assumed the rearrangment was some mysterious property of $e$. All these new numbers are scaring the sense out of me (Worried), although it was very late, I have so got to get a grip.

Thanks again:)