Can You Solve This Logarithmic Inequality?

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The discussion centers on solving a logarithmic inequality presented as an exam question. The original poster expresses difficulty in finding a solution and seeks assistance. Participants suggest applying logarithmic laws and changing the base of the logarithms to simplify the problem. They emphasize that while the inequality may seem daunting, using these strategies can lead to a solution. Overall, the conversation focuses on collaborative problem-solving techniques for logarithmic inequalities.
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Homework Statement


Here is the problem :
https://www.dropbox.com/s/otzzne7wjyuqa5o/question.jpg?dl=0
I've tried solving this inequality but alas,nothing...It's an exam question for my student,and for my great shame,I have no idea how to solve it :(
 
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What have you tried so far? I know it looks scary, but why don't you try using some of your laws of logarithms.
 
You might also want to use the fact that
log_a(x)= \frac{log_b(x)}{log_b(a)}
to get the two logarithms to the same base.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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