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$\log_2{\frac{x-3}{x+2}}≤0$

Thank you.

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closed as off-topic by T. Bongers, anorton, baba ji, Jack Lee, RecklessReckoner Jul 5 at 23:17

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is that log base 2? –  Vishwa Iyer Jul 5 at 21:20
3  
Can you share what you've tried, and explain what you're having trouble with? How do you approach inequalities like this in general? What previous examples have you seen, or what are some techniques you know? –  user61527 Jul 5 at 21:20
    
yeah it's log base 2;I tried solving this and finaly I got this result (-1/x+2)≤0 ,but I am really bad at this so anything will be helpful –  Heisenberg Jul 5 at 21:24
    
@Heisenberg Then please edit your question to include what you've tried - we can perhaps find any mistakes, or help you finish the reasoning. –  user61527 Jul 5 at 21:25
3  
@Heisenberg This isn't a do-my-homework-on-a-schedule site; if you just want the answer, then may I suggest Yahoo! Answers or WolframAlpha? –  user61527 Jul 5 at 21:30

3 Answers 3

up vote 3 down vote accepted

Hint:

1) Notice that the lefthand side only makes sense when $\frac{x-3}{x+2}>0$, since otherwise the logarithm is undefined.

2) Since $f(x)=2^x$ is increasing, we can raise 2 to both sides of the inequality to get $\frac{x-3}{x+2}\le1$.

The intersection of the solution sets for these two inequalities will give the answer.

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Hint: ask yourself the following question: when is a logarithm negative?

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HINT: Solve the equation as if the inequality didn't exist and it was just an equality.

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