Solving the system $\log_2(x^2y+\frac{xy^2}2)-\log_{1/2}(\frac1x+\frac2y)=3$, $\log_{1/5}\lvert\frac{xy}6\rvert=0$

Question

I trying ti finish the system of equations, but I don't know how to do that. Here is my solution so far:

$$\begin{cases} \log_{2} (x^2y + \frac{xy^2}{2}) - \log_{\frac{1}{2}}(\frac{1}{x} + \frac{2}{y}) = 3 \\ \log_{\frac{1}{5}} \lvert \frac{xy}{6} \rvert = 0 \end{cases}$$

$$\begin{cases} \frac{\ln (x^2y + \frac{xy^2}{2})}{\ln 2} - \frac{\ln(\frac{1}{x} + \frac{2}{y})}{\ln \frac{1}{2}} = 3 \\ \lvert \frac{xy}{6} \rvert = 1 \end{cases}$$

$$\begin{cases} \frac{\ln (x^2y + \frac{xy^2}{2})}{\ln 2} + \frac{\ln(\frac{1}{x} + \frac{2}{y})}{\ln 2} = 3 \\ \lvert \frac{xy}{6} \rvert = 1 \end{cases}$$

$$\begin{cases} \ln (x^2y + \frac{xy^2}{2}) + \ln(\frac{1}{x} + \frac{2}{y}) = 3 \ln 2 \\ \lvert \frac{xy}{6} \rvert = 1 \end{cases}$$

$$\begin{cases} \ln ((x^2y + \frac{xy^2}{2}) * (\frac{1}{x} + \frac{2}{y})) = 3 \ln 2 \\ \lvert \frac{xy}{6} \rvert = 1 \end{cases}$$

$$\begin{cases} (x^2y + \frac{xy^2}{2}) * (\frac{1}{x} + \frac{2}{y}) = 8 \\ \lvert \frac{xy}{6} \rvert = 1 \end{cases}$$

$$\begin{cases} 2xy + 2x^2 + \frac{y^2}{2} = 8 \\ \lvert \frac{xy}{6} \rvert = 1 \end{cases}$$

Answer 1

Note that$$2xy+2^2+\frac12y^2=2\left(x+\frac12y\right)^2.$$If $xy=6$, then$$2xy+2x^2+\frac12y^2=8\iff2x^2+\frac12y^2=-4,$$and this equation has no real solutions.

On the other hand, if $xy=-6$, then\begin{align}2xy+2x^2+\frac12y^2=8&\iff2\left(x+\frac12y\right)^2=8\\&\iff x+\frac12y=\pm2.\end{align}So, you solve the systems$$\left\{\begin{array}{l}x+\frac12y=\pm2\\xy=-6;\end{array}\right.$$you will get the solutions $(x,y)=\pm(-3,2)$ and $(x,y)=\pm(1,-6)$. Among these solutions, you take those pairs $(x,y)$ such that $x^2y+\frac12xy^2>0$ and $\frac1x+\frac2y>0$.