0
$\begingroup$

I am trying to solve this system of equations. I know the answer but I am struggling with the working. I need to solve for $m$ and $s^2$ in terms of all the other parameters. The system of equations is shown below

\begin{align} e^{(m+\frac{1}{2}s^2)}=l_1+l_2+(r-\frac{1}{2}\nu_1^2)t+(r-\frac{1}{2}\nu_2^2)t\qquad(1)\\ e^{(2m+s^2)}(e^{s^2}-1)=\nu_1^2t+\nu_2^2t+2h\qquad(2) \end{align}

I have noticed that the first term of the LHS of $(2)$ is the square of LHS $(1)$. Calling the RHS of $(1)$ and $(2)$ as $a$ and $b$ respectively, we have

\begin{align} e^{(2m+x)}&=a^2\\ a^2(e^x-1)&=b \end{align} where $s^{2}=x$ (say).

I have got to the point where,

\begin{align} m = 2\log a - \frac12 \log(b + a^2)\\ s^2 = \log(b+a^2) - 2\log a \end{align}

but struggling when doing the algebra, as in when replacing $a$ and $b$ by their corresponding terms in $(1)$ and $(2)$.

I would very much appreciate some help here.

$\endgroup$
0
$\begingroup$

Hint: Writing $$e^{m+\frac{1}{2}s^2}=a$$ and $$e^{2m+s^2}(e^{s^2}-1)=b$$ taking the logarithm on both sides and both equations we get $$m+\frac{1}{2}s^2=\ln(a)$$ $$2m+s^2+\ln(e^{s^2}-1)=\ln(b)$$ multiplying the fist equation by $$-2$$ and adding both we get $$\ln(e^{s^2}-1)=\ln(b)-2\ln(a)$$ from here you will get $$s$$ And then good luck!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy