The constant term in the expansion of `(log_(10)(x^(log_(10)x))-log_(x^(2))100)^(12)` is:

Options

A. 495
B. 924
C. 1050
D. 5050

Answer

To find the constant term in the expansion of \( \left( \log_{10}(x^{\log_{10} x}) - \log_{x^2} 100 \right)^{12} \), we will follow these steps:

### Step 1: Simplify the logarithmic expressions

1. **First term**:
\[ \log_{10}(x^{\log_{10} x}) = \log_{10} x \cdot \log_{10} x = (\log_{10} x)^2 \]

2. **Second term**:
\[ \log_{x^2} 100 = \frac{\log_{10} 100}{\log_{10} (x^2)} = \frac{2}{2 \log_{10} x} = \frac{1}{\log_{10} x} \]

Now, we can rewrite the expression inside the parentheses:
\[ \log_{10}(x^{\log_{10} x}) - \log_{x^2} 100 = (\log_{10} x)^2 - \frac{1}{\log_{10} x} \]

### Step 2: Combine the terms

Now, we can express the entire expression as:
\[ \left( (\log_{10} x)^2 - \frac{1}{\log_{10} x} \right)^{12} \]

### Step 3: General term in the binomial expansion

Let \( y = \log_{10} x \). The expression becomes:
\[ \left( y^2 - \frac{1}{y} \right)^{12} \]

Using the binomial theorem, the general term \( T_r \) in the expansion is given by:
\[ T_r = \binom{12}{r} (y^2)^{12-r} \left(-\frac{1}{y}\right)^r = \binom{12}{r} (-1)^r y^{24 - r} \]

### Step 4: Find the constant term

To find the constant term, we need the exponent of \( y \) to be zero:
\[ 24 - r = 0 \implies r = 24 \]

However, since \( r \) can only take values from 0 to 12, we need to adjust our approach. We will instead set:
\[ 24 - 3r = 0 \implies 3r = 24 \implies r = 8 \]

### Step 5: Calculate the constant term

Now, substituting \( r = 8 \) into the general term:
\[ T_8 = \binom{12}{8} (-1)^8 y^{24 - 3 \cdot 8} = \binom{12}{8} y^{0} = \binom{12}{8} \]

Calculating \( \binom{12}{8} \):
\[ \binom{12}{8} = \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11880}{24} = 495 \]

### Final Answer

Thus, the constant term in the expansion is \( \boxed{495} \).

---