, 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

$\text{( 10 x) 2 + 2( 10 x)( 10 y) + ( 10 y) 2 = 4 ( 10 4y) 2 + 2( 10 4y)( 10 y) + ( 10 y) 2 = 4 ( 10 4 + 10 y) 2 + 2( 10 4 + 10 y)( 10 y) + ( 10 y) 2 = 4 ( 10 4) 2 + 2( 10 y)( 10 4) + ( 10 y) 2 + 2( 10 y) 2 + 2( 10 y)( 10 4) + ( 10 y) 2 = 4 4( 10 y) 2 + 4( 10 y)( 10 4) + ( 10 4) 2 - 4 = 0}$$Let$a = $\log_{10}y$$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

$\text{( 10 x) 2 + 2( 10 x)( 10 y) + ( 10 y) 2 = 4 ( 10 4y) 2 + 2( 10 4y)( 10 y) + ( 10 y) 2 = 4 ( 10 4 + 10 y) 2 + 2( 10 4 + 10 y)( 10 y) + ( 10 y) 2 = 4 ( 10 4) 2 + 2( 10 y)( 10 4) + ( 10 y) 2 + 2( 10 y) 2 + 2( 10 y)( 10 4) + ( 10 y) 2 = 4 4( 10 y) 2 + 4( 10 y)( 10 4) + ( 10 4) 2 - 4 = 0}$$Let$c = $\log_{10}y$$and$d = $\log_{10}4$. We can rewrite the last equation above as

$\text{4c 2 + 4cd + d 2 - 4 = 0 4c 2 + 4cd + d 2 = 4 (2c+d) 2 = 4}$$and so$2c + d = -2$or$2c + d = 2$Thus,$$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 $(\frac{1}{5}, \frac{1}{20})$.