Manipulating An Algebraic Expression I'm trying to solve for $x$ in the following expression:
$$\frac{1}{s\sqrt{2\pi}}e^\frac{-(x-u)^2}{2s}=\frac{1}{w\sqrt{2\pi}}e^\frac{-(x-u)^2}{2w}$$
I'm able to rearrange it to this:
$$\frac{2 \log_e (\frac sw)(sw)^2}{s^2 - w^2} = (x-u)^2$$
but after taking the 2nd root of both sides I become stuck:
$$\sqrt\frac{2 \log_e (\frac sw)(sw)^2}{s^2 - w^2} = |x-u|$$
I don't know how to take the $u$ variable out of the absolute value. If anyone could tell me how you solve for x from the last expression I'd really appreciate it.
Thanks
 A: You have an expression of the form $(x-u)^2=a$ for some $a$. Then $x-u = \pm\sqrt a$ and so
$x = \pm\sqrt a +u$.
A: Well, you have an expression in the form:
$$\text{n}_1\exp\left(-\frac{\left(x-\alpha\right)^2}{\text{m}_1}\right)=\text{n}_2\exp\left(-\frac{\left(x-\alpha\right)^2}{\text{m}_2}\right)\tag1$$
Let's first divide:
$$\frac{\text{n}_1}{\text{n}_2}=\frac{\exp\left(-\frac{\left(x-\alpha\right)^2}{\text{m}_2}\right)}{\exp\left(-\frac{\left(x-\alpha\right)^2}{\text{m}_1}\right)}=\exp\left(-\frac{\left(x-\alpha\right)^2}{\text{m}_2}-\left(-\frac{\left(x-\alpha\right)^2}{\text{m}_1}\right)\right)=$$
$$\exp\left(\frac{\left(x-\alpha\right)^2}{\text{m}_1}-\frac{\left(x-\alpha\right)^2}{\text{m}_2}\right)=\exp\left(\left(x-\alpha\right)^2\cdot\left(\frac{1}{\text{m}_1}-\frac{1}{\text{m}_2}\right)\right)=$$
$$\exp\left(\left(x-\alpha\right)^2\right)\cdot\exp\left(\frac{1}{\text{m}_1}-\frac{1}{\text{m}_2}\right)\tag2$$

Can you conclude?


So:
$$\frac{\text{n}_1}{\text{n}_2}=\exp\left(\left(x-\alpha\right)^2\right)\cdot\exp\left(\frac{1}{\text{m}_1}-\frac{1}{\text{m}_2}\right)\space\Longleftrightarrow\space$$
$$\exp\left(\frac{1}{\text{m}_2}-\frac{1}{\text{m}_1}\right)\cdot\frac{\text{n}_1}{\text{n}_2}=\exp\left(\left(x-\alpha\right)^2\right)\space\Longleftrightarrow\space$$
$$\ln\left(\exp\left(\frac{1}{\text{m}_2}-\frac{1}{\text{m}_1}\right)\cdot\frac{\text{n}_1}{\text{n}_2}\right)=\left(x-\alpha\right)^2\space\Longleftrightarrow\space$$
$$\alpha\pm\sqrt{\frac{1}{\text{m}_2}-\frac{1}{\text{m}_1}+\ln\left(\text{n}_1\right)-\ln\left(\text{n}_2\right)}=x\tag3$$
