To determine the true solution(s) for the equation
\[ \log(4 + 2x) = \log(x^2 + 5x) \]
we can start by using the property of logarithms that states if \(\log(a) = \log(b)\), then \(a = b\) given that \(a > 0\) and \(b > 0\).
From the equation, we can set the arguments of the logarithms equal to each other:
\[ 4 + 2x = x^2 + 5x \]
Rearranging this equation gives:
\[ 0 = x^2 + 5x - 2x - 4 \]
which simplifies to:
\[ 0 = x^2 + 3x - 4 \]
Next, we can factor this quadratic:
\[ 0 = (x + 4)(x - 1) \]
Setting each factor to zero gives us the potential solutions:
\[ x + 4 = 0 \quad \text{or} \quad x - 1 = 0 \]
Thus, the solutions are:
\[ x = -4 \quad \text{and} \quad x = 1 \]
Now, we need to check if these solutions are valid by substituting them back into the original logarithmic expressions:
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For \(x = 1\):
- Left side: \(\log(4 + 2(1)) = \log(6)\)
- Right side: \(\log(1^2 + 5(1)) = \log(6)\)
- Since both sides are equal, \(x = 1\) is a valid solution.
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For \(x = -4\):
- Left side: \(\log(4 + 2(-4)) = \log(4 - 8) = \log(-4)\) (undefined)
- Right side: \(-4^2 + 5(-4) = 16 - 20 = -4\) (also undefined)
- Since the logarithm of a negative number is undefined, \(x = -4\) is not a valid solution.
Therefore, \(x = 1\) is the only true solution, and \(x = -4\) is an extraneous solution.
So the correct response is:
x = 1 is a true solution and x = −4 is an extraneous solution.