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Question

Question: $\begin{matrix} \log_3 512 & \log_2 3 & \log_8 3\\ \log_4 3 & \log_3 4 & \log_3 4\\ \log_3 8 & \log...

log3512log23log83log43log34log34log38log49\begin{matrix} \log_3 512 & \log_2 3 & \log_8 3\\ \log_4 3 & \log_3 4 & \log_3 4\\ \log_3 8 & \log_4 9 \end{matrix} is equal to

A

5

B

7

C

10

D

12

Answer

There appears to be an error in the question statement or the provided options, as the determinant does not simplify to a constant integer value based on standard calculations. Without clarification or correction of the question, it is not possible to definitively choose one of the integer options.

Explanation

Solution

The determinant is given by

Δ=log3512log23log83log43log34log34log38log49log364\Delta = \begin{vmatrix} \log_3 512 & \log_2 3 & \log_8 3\\ \log_4 3 & \log_3 4 & \log_3 4\\ \log_3 8 & \log_4 9 & \log_3 64 \end{vmatrix}

Using logarithm properties logban=nlogba\log_b a^n = n \log_b a, logbma=1mlogba\log_{b^m} a = \frac{1}{m} \log_b a, and logba=1logab\log_b a = \frac{1}{\log_a b}, we can rewrite the entries.

Let x=log32x = \log_3 2. Then log23=1log32=1x\log_2 3 = \frac{1}{\log_3 2} = \frac{1}{x}.

The entries become:

log3512=log329=9log32=9x\log_3 512 = \log_3 2^9 = 9 \log_3 2 = 9x

log23=1/x\log_2 3 = 1/x

log83=log233=13log23=13x\log_8 3 = \log_{2^3} 3 = \frac{1}{3} \log_2 3 = \frac{1}{3x}

log43=log223=12log23=12x\log_4 3 = \log_{2^2} 3 = \frac{1}{2} \log_2 3 = \frac{1}{2x}

log34=log322=2log32=2x\log_3 4 = \log_3 2^2 = 2 \log_3 2 = 2x

log38=log323=3log32=3x\log_3 8 = \log_3 2^3 = 3 \log_3 2 = 3x

log49=log2232=22log23=log23=1/x\log_4 9 = \log_{2^2} 3^2 = \frac{2}{2} \log_2 3 = \log_2 3 = 1/x

log364=log326=6log32=6x\log_3 64 = \log_3 2^6 = 6 \log_3 2 = 6x

The determinant is:

Δ=9x1/x1/(3x)1/(2x)2x2x3x1/x6x\Delta = \begin{vmatrix} 9x & 1/x & 1/(3x)\\ 1/(2x) & 2x & 2x\\ 3x & 1/x & 6x \end{vmatrix}

Expanding the determinant:

Δ=9x[(2x)(6x)(2x)(1/x)]1/x[(1/(2x))(6x)(2x)(3x)]+1/(3x)[(1/(2x))(1/x)(2x)(3x)]\Delta = 9x [(2x)(6x) - (2x)(1/x)] - 1/x [(1/(2x))(6x) - (2x)(3x)] + 1/(3x) [(1/(2x))(1/x) - (2x)(3x)]

Δ=9x[12x22]1/x[36x2]+1/(3x)[1/(2x2)6x2]\Delta = 9x [12x^2 - 2] - 1/x [3 - 6x^2] + 1/(3x) [1/(2x^2) - 6x^2]

Δ=(108x318x)(3/x6x)+(1/(6x3)2x)\Delta = (108x^3 - 18x) - (3/x - 6x) + (1/(6x^3) - 2x)

Δ=108x318x3/x+6x+1/(6x3)2x\Delta = 108x^3 - 18x - 3/x + 6x + 1/(6x^3) - 2x

Δ=108x314x3/x+1/(6x3)\Delta = 108x^3 - 14x - 3/x + 1/(6x^3)

Where x=log32x = \log_3 2. This expression is generally not an integer and depends on the value of xx. Given that the options are integers, there is likely an error in the question statement or the provided options, as the determinant does not simplify to a constant integer value based on standard calculations. Without clarification or correction of the question, it is not possible to definitively choose one of the integer options.