To solve the equation \( \log(2x^2 - 10) = \log(x^2 - 3x) \), we can use the property of logarithms that states if \( \log(a) = \log(b) \), then \( a = b \) (as long as \( a > 0 \) and \( b > 0 \)).
So we will set the arguments equal to each other:
\[ 2x^2 - 10 = x^2 - 3x \]
Now, let's rearrange the equation:
\[ 2x^2 - x^2 + 3x - 10 = 0 \]
This simplifies to:
\[ x^2 + 3x - 10 = 0 \]
Next, we can factor the quadratic equation:
\[ (x + 5)(x - 2) = 0 \]
Setting each factor to zero gives us the solutions:
\[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
So, the potential solutions are \( x = -5 \) and \( x = 2 \).
Now we need to check if these solutions are valid by substituting them back into the original logarithmic expression to ensure that we do not take the log of a non-positive number.
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For \( x = 2 \): \[ 2x^2 - 10 = 2(2^2) - 10 = 8 - 10 = -2 \quad \text{(not valid)} \]
\[ x^2 - 3x = 2^2 - 3(2) = 4 - 6 = -2 \quad \text{(not valid)} \]
So, \( x = 2 \) is not a valid solution.
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For \( x = -5 \): \[ 2x^2 - 10 = 2(-5)^2 - 10 = 50 - 10 = 40 \quad \text{(valid)} \]
\[ x^2 - 3x = (-5)^2 - 3(-5) = 25 + 15 = 40 \quad \text{(valid)} \]
Since \( x = -5 \) is valid while \( x = 2 \) is not valid, the only solution is:
\[ \boxed{-5} \]
The correct response is: x is equal to negative 5.