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In solving this inequality (transcribed from here)

$$\left(\frac23\right)^{\log_{0.5}(x^2+4x+4)}<\left(\frac94\right)^{\log_2(x^2-3x-10)}$$

we eventually reach the point where $ \frac{x^2-10x+24}{(x+2)^2(x-5)^2}>0 $ and later solve $x^2-10x+24>0$ to obtain a certain solution which we later need to intersect with a definition range.

Question is, whether definition range of $\frac{x^2-10x+24}{(x+2)^2(x-5)^2}$ is same like that of the original inequality or do we have to include its conditions in the total eventual definition range separately (add more conditions to the initial ones), to give us the complete range of definition?

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Note : log{ base 1/2} w = - log {base 2 } w

(2/3) ^(log { base 1/2} (x+2)² ) = (3/2)^(log {base 2} (x+2)² ) = (9/4)^(log {base 2 } (x+2) )

< (9/4)^log{base 2 } (x² - 3x -10) means (x + 2 ) < (x² - 3x - 10 ) OR

0 < x² - 4x -12) = ( x - 6 )(x+2)---> x in (-∞ , - 2 ) U (6 , ∞)

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Assuming your calculations to be correct,

we need to have $x^2+4x+4=(x+2)^2, x^2-3x-10=(x-5)(x+2)>0$

The first one is already perfect square, so only needs $x+2\ne0\iff x\ne-2$

The second needs $x>$max$(5,-2)=5$ or $x<$min$(5,-2)$

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