To solve the inequality \( \log_{0.75}(x + 3) > \log_{0.5}(x + 3) \), we can first convert the logarithms to the same base, or we can analyze the properties of the logarithmic functions involved.
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Understanding the log properties: The base 0.75 logarithm is decreasing because 0.75 is less than 1, while the base 0.5 logarithm is also decreasing for the same reason. Thus, for any valid input \( x + 3 > 0 \), we can compare \( x + 3 \) directly.
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Reformulating the inequality: Since both logarithms are decreasing functions, we can reverse the inequality when we take the argument of the logarithms:
\[ x + 3 < (x + 3) \]
However, this does not hold because we're comparing two functions rather than two quantities being directly compared.
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Using the change of base: We can express the logarithm conversion:
\[ \log_{0.75}(x + 3) = \frac{\log(x + 3)}{\log(0.75)} \] \[ \log_{0.5}(x + 3) = \frac{\log(x + 3)}{\log(0.5)} \]
Thus, our inequality becomes:
\[ \frac{\log(x + 3)}{\log(0.75)} > \frac{\log(x + 3)}{\log(0.5)} \]
Since \( \log(0.75) < 0 \) and \( \log(0.5) < 0 \), we multiply both sides by the negative \( \log(0.75) \) (which flips the inequality):
\[ \log(x + 3) < \log(x + 3) \cdot \frac{\log(0.5)}{\log(0.75)} \]
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Finding the condition: Recall that an inequality of logarithmic forms indicates the relationship of the arguments. We need to validate:
\[ (x + 3)^{\frac{\log(0.75)}{\log(0.5)}} < 1 \]
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Determine when this occurs: Since \(\log(0.75) < 0\) and \(\log(0.5) < 0\),
\[ \frac{\log(0.75)}{\log(0.5)} > 1 \]
Meaning:
\[ x + 3 < 1 \] Simplifying gives:
\[ x < -2 \]
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Consider constraints: Since the logarithmic function is defined for positive arguments, we must also have:
\[ x + 3 > 0 \implies x > -3 \]
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Final interval: Now, we need to satisfy both conditions: \(x < -2\) and \(x > -3\). The solution is the interval:
\[ -3 < x < -2 \]
So, the solution for the inequality \( \log_{0.75}(x + 3) > \log_{0.5}(x + 3) \) is:
\[ \boxed{(-3, -2)} \]