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