# Homework Help: Sigh, I don't understand why I am not getting a good solution

1. Aug 3, 2011

### flyingpig

1. The problem statement, all variables and given/known data

Solve for all x that satisifies

$$log_{0.5} \frac{2x - 5}{x + 2} < 0$$

3. The attempt at a solution

Let me just say that upon doing this problem i learned that

1) When you exponeitate something to the base in (0, 1), you flip the inequality sign.

So I finally got the solution to be (-infy,-2) U (7, infty)

Now i found that this approach does not work

$$log_{0.5} \left | \frac{2x - 5}{x + 2} \right | < 0$$

SO it would make sense that

$$\left | \frac{2x - 5}{x + 2} \right | < 0$$

Okay after doing some test points for

$$\frac{2x - 5}{x + 2} < 0$$

$$\frac{5 - 2x}{x + 2} > 0$$

I found that there are no solution to the absolute value ...

Last edited: Aug 3, 2011
2. Aug 3, 2011

### BruceW

You did 0.5^(your equation) for both sides, so this does get:
$$\frac{2x-5}{x+2}$$
for the left-hand side, but for the right-hand side, what is 0.5^0 ?

3. Aug 3, 2011

### flyingpig

No i already solved it using that method but I am asking why I am wrong in the other.

4. Aug 3, 2011

### flyingpig

Oh shoot, I meant

$$\left | \frac{2x - 5}{x + 2} \right | > 0$$

5. Aug 4, 2011

### vela

Staff Emeritus
[strike]It's because$$\log_{1/2} \left\lvert \frac{2x-5}{x+2} \right\rvert < 0$$ doesn't imply $$\left\lvert\frac{2x-5}{x+2} \right\rvert > 0$$Why would you think it does?[/strike]

Last edited: Aug 4, 2011
6. Aug 4, 2011

### flyingpig

Because you can't take a log of a negative number

That's why I introduce a new inequality?

7. Aug 4, 2011

### vela

Staff Emeritus
Sorry, you're right. $$\log_{1/2} \left\lvert \frac{2x-5}{x+2} \right\rvert < 0$$ does imply $$\left\lvert\frac{2x-5}{x+2} \right\rvert > 0$$The problem is the implication doesn't work the other way. There are values of x for which$$\left\lvert\frac{2x-5}{x+2} \right\rvert > 0$$ and $$\log_{1/2} \left\lvert \frac{2x-5}{x+2} \right\rvert > 0$$
This is like the fact that x>2 implies x2>4, but x2>4 doesn't imply x>2 because the inequality x2>4 admits more solutions than x>2. In the same way,$$\left\lvert\frac{2x-5}{x+2} \right\rvert > 0$$ is satisfied by more values of x than $$\log_{1/2} \left\lvert \frac{2x-5}{x+2} \right\rvert < 0$$

8. Aug 4, 2011

### flyingpig

But you couldn't take the log of a negative number int he first place with $$\log_{1/2} \left\lvert \frac{2x-5}{x+2} \right\rvert > 0$$

Is there any meaning to creating a new inequality

$$\left\lvert\frac{2x-5}{x+2} \right\rvert > 0$$ then?

$$log_{0.5} \left | \frac{2x - 5}{x + 2} \right | = log_{0.5} (2x - 5) - log_{0.5} (x + 2)$$

Then it is only true for (5/2, infintiy), so I destroyed the solution in (-infinty, -2)

Last edited: Aug 4, 2011
9. Aug 4, 2011

### vela

Staff Emeritus
Not really. So why are you doing it?
Your mistake is in removing the absolute values from the righthand side. The two sides of your "equation" are equal only when 2x-5>0 and x+2>0.

10. Aug 4, 2011

### Bohrok

May I ask why you're introducing absolute values around the fraction in the log? I can't see the justification in using them, plus you don't need them to solve rational inequalities like this one.

11. Aug 4, 2011

### BruceW

$$log_{0.5} \left | \frac{2x - 5}{x + 2} \right | = log_{0.5} (2x - 5) - log_{0.5} (x + 2)$$
And then he found there were problems when x=-3 (for example), the right-hand side of the equation goes bad (since we're talking about real numbers here).
So vela is explaining that you need to introduce the absolute values, or this rearrangement doesn't hold.

12. Aug 4, 2011

### vela

Staff Emeritus
I think Bohrok may have been asking why Flyingpig introduced the absolute value in the first place. I didn't notice that the original problem didn't have them.

13. Aug 4, 2011

### SammyS

Staff Emeritus
Basic stuff (for my friend, flyingpig):

For real numbers:
1. The expression $\log_{\,a}(x)$ is defined only for x > 0 .
Making sure that you're taking the log of a positive quantity should be part of working out your solution.

2. If a > 1, then $\log_{\,a}(x)<0$ implies that x < 1 .

3. If 0 < a < 1, then $\log_{\,a}(x)<0$ implies that x > 1 .

14. Aug 5, 2011

### flyingpig

Sammy, so for #3, I think that |-x| also works?

15. Aug 5, 2011

### Staff: Mentor

Assuming that you are referring to #3 in Sammy's list, why would you introduce this extra complication? Whatever x's value (assumed real), |-x| = |x| and is a nonnegative number.