Suppose `a,b,cgt1`. Let
`Delta=|(loga,logb,logc),(log(2007a),log(2007b),log(2007c)),(log(2017a),log(2017b),log(2017c))|` then `Delta` is equal to

Options

A. 0
B. `log(4024abc)`
C. `log(2017/2007)`
D. None of these

Answer

To solve the determinant \( \Delta = \begin{vmatrix} \log a & \log b & \log c \\ \log(2007a) & \log(2007b) & \log(2007c) \\ \log(2017a) & \log(2017b) & \log(2017c) \end{vmatrix} \), we will follow these steps:

### Step 1: Expand the logarithms
We can express the logarithms of products using the property \( \log(xy) = \log x + \log y \).

\[ \log(2007a) = \log 2007 + \log a, \quad \log(2007b) = \log 2007 + \log b, \quad \log(2007c) = \log 2007 + \log c \]
\[ \log(2017a) = \log 2017 + \log a, \quad \log(2017b) = \log 2017 + \log b, \quad \log(2017c) = \log 2017 + \log c \]

### Step 2: Rewrite the determinant
Substituting these expressions into the determinant, we have:

\[ \Delta = \begin{vmatrix} \log a & \log b & \log c \\ \log 2007 + \log a & \log 2007 + \log b & \log 2007 + \log c \\ \log 2017 + \log a & \log 2017 + \log b & \log 2017 + \log c \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out the common terms from the second and third rows:

\[ \Delta = \begin{vmatrix} \log a & \log b & \log c \\ \log a & \log b & \log c \\ \log a & \log b & \log c \end{vmatrix} + \log 2007 \begin{vmatrix} \log a & \log b & \log c \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{vmatrix} + \log 2017 \begin{vmatrix} \log a & \log b & \log c \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{vmatrix} \]

### Step 4: Identify rows that are the same
Notice that the second and third rows of the determinant are identical, which means the determinant evaluates to zero:

\[ \Delta = 0 \]

### Conclusion
Thus, the value of \( \Delta \) is:

\[ \Delta = 0 \]