CEMC Euclid, 2025

Let $b = g^{-1}(a)$. Since
$f^{-1}(b) = 3$, then $b = f(3) = 4$.

Since $g^{-1}(a) = b = 4$, then $a= g(4) = 5$.
Solution 1:

The first equation can be rewritten as $(x - 4y)^2 = 0$, from which we obtain $x - 4y = 0$ or $x = 4y$.

The second equation can be rewritten as $(\log_{10}x + \log_{10}y)^2 = 4$, from
which we obtain $\log_{10}x + \log_{10}y = \pm 2$.

Using logarithm rules, $\log_{10}(xy) = \pm 2$ and so $xy = 10^2 = 100$
or $xy = 10^{-2} = \frac{1}{100}$.

Since $x = 4y$, then $4y^2 = 100$ or $4y^2 = \frac{1}{100}$, which gives $y^2 = 25$ or $y^2 = \frac{1}{400}$.

Since $\text{y gt; 0}$ (because of the
domain of a logarithm), then $y = 5$
or $y = \frac{1}{20}$.

Since $x = 4y$, then $x = 20$ or $x = \frac{1}{5}$.

Therefore, $(x,y) = (20, 5)$ or $(\frac{1}{5}, \frac{1}{20})$.

Solution 2:

The first equation can be rewritten as $(x - 4y)^2 = 0$, from which we obtain $x - 4y = 0$ or $x = 4y$.

The second equation can thus be rewritten successively as $$\begin{align*} (\log_{10}x)^2 + 2(\log_{10}x)(\log_{10}y) + (\log_{10}y)^2 & = 4\\ (\log_{10}4y)^2 + 2(\log_{10}4y)(\log_{10}y) + (\log_{10}y)^2 & = 4\\ (\log_{10}4 + \log_{10}y)^2 + 2(\log_{10}4 + \log_{10}y)(\log_{10}y) + (\log_{10}y)^2 & = 4\\ (\log_{10}4)^2 + 2(\log_{10}y)(\log_{10}4) + (\log_{10}y)^2 + 2(\log_{10}y)^2 + 2(\log_{10}y)(\log_{10}4) + (\log_{10}y)^2 & = 4\\ 4(\log_{10}y)^2 + 4(\log_{10}y)(\log_{10}4) + (\log_{10}4)^2 - 4 & = 0 \\\end{align*}$$ Let $a = \log_{10}y$ and $b = \log_{10}2$. Then $2b = 2\log_{10}2 = \log_{10}2^2 = \log_{10}4$.

We can rewrite the last equation above as $$\begin{align*} 4a^2 + 8ab + 4b^2 - 4 & = 0 \\ a^2 + 2ab +b^2 & = 1\\ (a+b)^2 & = 1 \end{align*}$$ and so $a + b = -1$ or $a + b = 1$

Thus, $\log_{10}y + \log_{10}2 = -1$
or $\log_{10}y + \log_{10}2 = 1$,
which simplify to give $\log_{10}2y = -1$ or $\log_{10}2y = 1$.

This means that $2y = \frac{1}{10}$
or $2y = 10$, and so $y = \frac{1}{20}$ or $y = 5$.

Since $x = 4y$, then $(x,y) = (20, 5)$ or $(\frac{1}{5}, \frac{1}{20})$.

Solution 3:

The first equation can be rewritten as $(x - 4y)^2 = 0$, from which we obtain $x - 4y = 0$ or $x = 4y$.

The second equation can thus be rewritten successively as $$\begin{align*} (\log_{10}x)^2 + 2(\log_{10}x)(\log_{10}y) + (\log_{10}y)^2 & = 4\\ (\log_{10}4y)^2 + 2(\log_{10}4y)(\log_{10}y) + (\log_{10}y)^2 & = 4\\ (\log_{10}4 + \log_{10}y)^2 + 2(\log_{10}4 + \log_{10}y)(\log_{10}y) + (\log_{10}y)^2 & = 4\\ (\log_{10}4)^2 + 2(\log_{10}y)(\log_{10}4) + (\log_{10}y)^2 + 2(\log_{10}y)^2 + 2(\log_{10}y)(\log_{10}4) + (\log_{10}y)^2 & = 4\\ 4(\log_{10}y)^2 + 4(\log_{10}y)(\log_{10}4) + (\log_{10}4)^2 - 4 & = 0 \\\end{align*}$$ Let $c = \log_{10}y$ and $d = \log_{10}4$. We can rewrite the last equation above as $$\begin{align*} 4c^2 + 4cd + d^2 - 4 & = 0 \\ 4c^2 + 4cd + d^2 & = 4 \\ (2c+d)^2 & = 4\end{align*}$$ and so $2c + d = -2$ or $2c + d = 2$

Thus, $2\log_{10}y + \log_{10}4 = -2$
or $2\log_{10}y + \log_{10}4 = 2$.

These simplify to give $\log_{10}(4y^2) = -2$ or $\log_{10}(4y^2) = 2$.

This means that $4y^2 = \frac{1}{100}$ or $4y^2 = 100$, and so $y = \pm \frac{1}{20}$ or $y = \pm 5$.

Since $\text{y gt; 0}$ and $x = 4y$, then $(x,y) = (20, 5)$ or $(\frac{1}{5}, \frac{1}{20})$.