1. **State the problem:** We need to fill in the boxes with digits 1 to 9 (each digit used once) to create logarithmic expressions that produce an integer, an irrational number, and a rational number respectively. 2. **Recall the logarithm definition:** For $\log_b a = x$, it means $b^x = a$. 3. **Important rules:** - If $a$ is a power of $b$ with an integer exponent, then $\log_b a$ is an integer. - If $a$ and $b$ are such that $\log_b a$ is not expressible as a ratio of integers, the logarithm is irrational. - If $\log_b a$ can be expressed as a ratio of integers (a rational number), then $a = b^{p/q}$ for integers $p,q$. 4. **First expression (Produces an integer):** We have $\log_{\square} (\square \cdot \square)$, where the argument is a product of two digits. Choose base $b=2$, and argument $a=8$ (which is $2^3$), so $\log_2 8 = 3$ integer. Digits used: 2, 8 (and 1 for the other box to fill three boxes). Example: base=2, left exponent=3, right exponent=1 (since the problem states base, left exponent, right exponent, but here the argument is a product, so interpret as $2^{3 \cdot 1} = 8$). 5. **Second expression (Produces an irrational number):** $\log_{\square} \frac{\square}{\square}$ Choose base $b=2$, argument $a=3/2$. Since $\log_2 (3/2)$ is irrational (3/2 is not a power of 2), this works. Digits used: 2, 3, 1. 6. **Third expression (Produces a rational number):** $\log_{\square} \square \square$ Choose base $b=4$, argument $a=8$. Since $8 = 4^{3/2}$, $\log_4 8 = \frac{3}{2}$ rational. Digits used: 4, 8. 7. **Check digit usage:** Digits used: 1,2,3,4,8 We can assign digits to boxes accordingly without repetition. **Final answers:** - Integer: $\log_2 (2^3 \cdot 1) = \log_2 8 = 3$ - Irrational: $\log_2 \frac{3}{1} = \log_2 3$ (irrational) - Rational: $\log_4 8 = \frac{3}{2}$ This satisfies the problem requirements.