Question

Simplify: $\left|\begin{array}{cc}\log_{2}512 & \log_{4}3 \\ \log_{8} & \log_{4}9\end{array}\right| \times \left|\begin{array}{cc}\log_{3}3 & \log_{2}3 \\ \log_{4} & \log_{2}4\end{array}\right|$

Answer 1

The question is incomplete due to missing arguments in the logarithm terms ($\log_8$ and $\log_4$).

Answer 2

The question as presented contains missing arguments for two logarithm terms: $\log_{8}$ in the first determinant and $\log_{4}$ in the second determinant. Without these arguments, the values of these terms cannot be determined, making the problem unsolvable.

Assuming this is a typo and the question intended to have complete entries, we analyze the most probable interpretations:

Interpretation 1: The missing arguments are 1 (i.e., $\log_b 1 = 0$). In this case, $\log_8$ is interpreted as $\log_8 1 = 0$, and $\log_4$ is interpreted as $\log_4 1 = 0$.

Let's evaluate the known terms first:

• $\log_2 512 = \log_2 2^9 = 9$
• $\log_4 3 = \frac{\log_2 3}{\log_2 4} = \frac{\log_2 3}{2}$
• $\log_4 9 = \log_4 3^2 = 2 \log_4 3 = 2 \times \frac{\log_2 3}{2} = \log_2 3$
• $\log_3 3 = 1$
• $\log_2 3$ (remains as is)
• $\log_2 4 = \log_2 2^2 = 2$

Under Interpretation 1, the two determinants are: First determinant, $D_1$: $D_1 = \left|\begin{array}{cc}\log_{2}512 & \log_{4}3 \\ \log_{8}1 & \log_{4}9\end{array}\right| = \left|\begin{array}{cc}9 & \frac{1}{2}\log_{2}3 \\ 0 & \log_{2}3\end{array}\right|$ Calculating $D_1$: $D_1 = (9)(\log_2 3) - (0)\left(\frac{1}{2}\log_2 3\right) = 9 \log_2 3$

Second determinant, $D_2$: $D_2 = \left|\begin{array}{cc}\log_{3}3 & \log_{2}3 \\ \log_{4}1 & \log_{2}4\end{array}\right| = \left|\begin{array}{cc}1 & \log_{2}3 \\ 0 & 2\end{array}\right|$ Calculating $D_2$: $D_2 = (1)(2) - (0)(\log_2 3) = 2$

The product $D_1 \times D_2 = (9 \log_2 3) \times (2) = 18 \log_2 3$.

Interpretation 2: The missing arguments are the base itself (i.e., $\log_b b = 1$). In this case, $\log_8$ is interpreted as $\log_8 8 = 1$, and $\log_4$ is interpreted as $\log_4 4 = 1$.

Under Interpretation 2, the two determinants are: First determinant, $D_1$: $D_1 = \left|\begin{array}{cc}\log_{2}512 & \log_{4}3 \\ \log_{8}8 & \log_{4}9\end{array}\right| = \left|\begin{array}{cc}9 & \frac{1}{2}\log_{2}3 \\ 1 & \log_{2}3\end{array}\right|$ Calculating $D_1$: $D_1 = (9)(\log_2 3) - (1)\left(\frac{1}{2}\log_2 3\right) = 9 \log_2 3 - \frac{1}{2}\log_2 3 = \left(9 - \frac{1}{2}\right)\log_2 3 = \frac{17}{2}\log_2 3$

Second determinant, $D_2$: $D_2 = \left|\begin{array}{cc}\log_{3}3 & \log_{2}3 \\ \log_{4}4 & \log_{2}4\end{array}\right| = \left|\begin{array}{cc}1 & \log_{2}3 \\ 1 & 2\end{array}\right|$ Calculating $D_2$: $D_2 = (1)(2) - (1)(\log_2 3) = 2 - \log_2 3$

The product $D_1 \times D_2 = \left(\frac{17}{2}\log_2 3\right) \times (2 - \log_2 3) = 17 \log_2 3 - \frac{17}{2}(\log_2 3)^2$.

Both interpretations lead to expressions involving $\log_2 3$, which cannot be simplified further to a numerical value unless $\log_2 3$ itself is a specific value or cancels out. Given the typical nature of "simplify" problems in exams, a simple numerical answer is often expected. Neither of these interpretations yields a simple numerical answer.

Therefore, the question is fundamentally incomplete due to the missing arguments in the logarithm terms. It is impossible to provide a definitive simplified answer without this information.

If forced to choose the most common convention for such a typo (or if it's a fill-in-the-blank question where the simplest expression is sought), Interpretation 1 leads to a slightly simpler expression ($18 \log_2 3$) compared to Interpretation 2 ($17 \log_2 3 - \frac{17}{2}(\log_2 3)^2$). However, without context or options, stating the question is incomplete is the most accurate response.

The problem statement has missing values. Assuming the missing values are $X$ and $Y$: $D_1 = \left|\begin{array}{cc}\log_{2}512 & \log_{4}3 \\ \log_{8}X & \log_{4}9\end{array}\right|$ $D_2 = \left|\begin{array}{cc}\log_{3}3 & \log_{2}3 \\ \log_{4}Y & \log_{2}4\end{array}\right|$

$D_1 = \left|\begin{array}{cc}9 & \frac{1}{2}\log_{2}3 \\ \frac{\log_2 X}{3} & \log_{2}3\end{array}\right| = 9\log_2 3 - \frac{1}{2}\log_2 3 \cdot \frac{\log_2 X}{3} = \log_2 3 \left(9 - \frac{\log_2 X}{6}\right)$ $D_2 = \left|\begin{array}{cc}1 & \log_{2}3 \\ \frac{\log_2 Y}{2} & 2\end{array}\right| = 2 - \log_2 3 \cdot \frac{\log_2 Y}{2} = 2 - \frac{\log_2 3 \cdot \log_2 Y}{2}$

The product is $\log_2 3 \left(9 - \frac{\log_2 X}{6}\right) \left(2 - \frac{\log_2 3 \cdot \log_2 Y}{2}\right)$. This cannot be simplified without $X$ and $Y$.