Question

Sole for x and y
e^(2x) + e^(2y) = 10
e^(x+y) = 4
hint: Let p=e^x and q=e^y
thanks

Answer 1

Substituting \(p=e^x\) and \(q=e^y\), you have
\[\begin{cases}p^2+q^2=10\\pq=4\end{cases}\]
The second equation is due to the fact that \(e^{x+y}=e^xe^y=pq\), in case you were wondering.
First, solve for either \(p\) or \(q\) in the second equation. I'll work with \(p\):
\[p=\frac{4}{q}\]
Next, plug this into the first equation:
\[\left(\frac{4}{q}\right)^2+q^2=10\\
\frac{16}{q^2}+q^2=10\\
16+q^4=10q^2\\
q^4-10q^2+16=0\\
(q^2-8)(q^2-2)=0\\
(q-\sqrt8)(q+\sqrt8)(q-\sqrt2)(q+\sqrt2)=0\]
So, \(q=\pm\sqrt8,~\pm\sqrt2\).
Now solve for \(p\).