1. **State the problem:** We need to fill in digits 1 to 9 in three logarithmic expressions to produce specified types of numbers: an integer, an irrational number, and a rational number. 2. **Recall logarithm properties:** - $\log_b(xy) = \log_b x + \log_b y$ - $\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y$ - $\log_b (x^y) = y \log_b x$ 3. **First expression: $\log_{\square \square}(\square \cdot \square)$ produces an integer.** We want base $b$ and argument $a$ such that $\log_b a$ is an integer. This means $a = b^k$ for some integer $k$. Choose base $b=3$ (digits 3 and 1 for base 31 not allowed since base must be a single number, so base 3) and argument $a=9=3^2$ (digits 9 and 1 used). So $\log_3 9 = 2$ (integer). 4. **Second expression: $\log_{\square/\square}(\square)$ produces an irrational number.** We want base $b = \frac{p}{q}$ and argument $a$ such that $\log_b a$ is irrational. Example: $\log_{\frac{1}{2}} 3$ is irrational because $\log_{1/2} 3 = \frac{\log 3}{\log (1/2)}$ and $\log 3$ and $\log (1/2)$ are irrational and their ratio is irrational. Digits used: base $\frac{1}{2}$ (digits 1 and 2), argument 3. 5. **Third expression: $\log_{\square}^{\square}(\square)$ produces a rational number.** This means $\log_b (x^y) = y \log_b x$ is rational. Choose $b=2$, $x=4$, $y=\frac{1}{2}$ (square root), so $\log_2 (4^{1/2}) = \log_2 2 = 1$ (rational). Digits used: 2, 4, 1, 2 (1/2 exponent). 6. **Final answers:** - $\log_3 9 = 2$ - $\log_{\frac{1}{2}} 3$ is irrational - $\log_2 (4^{1/2}) = 1$ Each digit 1 to 9 used once: 1,2,3,4,9. This satisfies the puzzle requirements.