Number of real values of x satisfying the equation `log_(x^2+6x+8)(log_(2x^2+2x+3)(x^2-2x))=0` is equal to

Explanation

To solve the equation \( \log_{(x^2 + 6x + 8)} \left( \log_{(2x^2 + 2x + 3)} (x^2 - 2x) \right) = 0 \), we will follow these steps:

### Step 1: Understand the logarithmic equation
The equation \( \log_{(x^2 + 6x + 8)} \left( \log_{(2x^2 + 2x + 3)} (x^2 - 2x) \right) = 0 \) implies that:
\[ \log_{(2x^2 + 2x + 3)} (x^2 - 2x) = 1 \]
This is because \( \log_b(a) = 0 \) means \( a = b^0 = 1 \).
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