1. Help With Rearranging!

Hey everyone!

I have the following line which needs to be expressed for x in terms of t:

$t = ln|1-x| - ln|2-x| + c$

What I first think of doing is raising everything as a power of e. However, this causes the x's to cancel.

eg. $e^t = 1-x-(2-x) +e^c$

As you can see the x's cancel and I am stuck.

Any help would be awesome!

Cheers.

2. Originally Posted by falconed
Hey everyone!

I have the following line which needs to be expressed for x in terms of t:

$t = ln|1-x| - ln|2-x| + c$

What I first think of doing is raising everything as a power of e. However, this causes the x's to cancel.

eg. $e^t = 1-x-(2-x) +e^c$

As you can see the x's cancel and I am stuck.

Any help would be awesome!

Cheers.
$t = \ln|1-x| - \ln|2-x| + c = \ln \left| \frac{1-x}{2-x} \right| + c$

$\Rightarrow t - c = \ln \left| \frac{1-x}{2-x} \right|$

$\Rightarrow e^{t - c} = \frac{1-x}{2-x}$

$\Rightarrow A e^t = \frac{1-x}{2-x}$ where $A = e^{-c}$ is just as arbitrary as c

$\Rightarrow A (2 - x) e^t = 1 - x$

$\Rightarrow 2A e^t - A x e^t = 1 - x$

$\Rightarrow x\left( 1 - A e^t \right) = 1 - 2A e^t$

$\Rightarrow x = \frac{1 - 2A e^t}{1 - A e^t}$.

No doubt there are many careless mistakes in the above so you should check it very carefully.

3. Originally Posted by mr fantastic
$t = \ln|1-x| - \ln|2-x| + c = \ln \left| \frac{1-x}{2-x} \right| + c$

$\Rightarrow t - c = \ln \left| \frac{1-x}{2-x} \right|$

$\Rightarrow e^{t - c} = \frac{1-x}{2-x}$

$\Rightarrow A e^t = \frac{1-x}{2-x}$ where $A = e^{-c}$ is just as arbitrary as c

$\Rightarrow A (2 - x) e^t = 1 - x$

$\Rightarrow 2A e^t - A x e^t = 1 - x$

$\Rightarrow x\left( 1 - A e^t \right) = 1 - 2A e^t$

$\Rightarrow x = \frac{1 - 2A e^t}{1 - A e^t}$.

No doubt there are many careless mistakes in the above so you should check it very carefully.

Thanks! I completely forgot that:

$ln |a| - ln |b| = ln |a/b|$

Cheers mate.