To solve the inequality \( \log_{0.75}(x + 3) > \log_{0.5}(x + 3) \), we can utilize the properties of logarithms.
First, we'll rearrange the logarithmic terms. Remember that the logs are defined for \( x + 3 > 0 \), which implies \( x > -3 \).
Next, let's analyze the logs:
- Since \(0.75 < 1\), the logarithm \( \log_{0.75}(x + 3) \) is a decreasing function.
- Since \(0.5 < 1\), the logarithm \( \log_{0.5}(x + 3) \) is also a decreasing function.
Thus, if \( \log_{0.75}(x + 3) > \log_{0.5}(x + 3) \), it implies that \( x + 3 < (x + 3) \) corresponds to values where \( x + 3 \) is in the range of values such that the logarithm from the smaller base (0.75) will yield a result greater than the logarithm from the larger base (0.5).
To solve the inequality, we can apply the change of base formula. Recall that for bases less than 1, the logarithmic function decreases:
\[ \log_{0.75}(x + 3) - \log_{0.5}(x + 3) > 0 \]
To resolve this, we can equate:
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Rewrite this as:
\[ \frac{\log(x + 3)}{\log(0.75)} < \frac{\log(x + 3)}{\log(0.5)} \]
This is simplified as \( \log(x + 3) \) must be negative given \( \log(0.75) < \log(0.5) \) because both expressions become negative for positive arguments of the logarithm.
Thus:
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We can also evaluate where exactly \( x + 3 \) holds:
- Since we are comparing, we evaluate :
\[ \log_{0.75} > \log_{0.5} \Rightarrow \frac{-\log(0.75)}{-\log(0.5)} < 1 \]
For further equality points, we must also note we can solve through testing.
From the inequality's behavior, we find \( x + 3 \) must fall significantly between bounds:
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Validating through testing values say around :
- \( x = -2\) yields:
We deduce it is significant to find ranges around overlaps.
The final result yields that the valid solution will be:
\[ \text{Negative 3} < x < \text{Negative 2} \]
Thus the correct response is:
Negative 3 less than x less than negative 2.