Sigh, I don't understand why I am not getting a good solution

[flyingpig]

Homework Statement

Solve for all x that satisifies

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

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 ...

 

[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 ?

 

[flyingpig]

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

 

[flyingpig]

Oh shoot, I meant

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

 

[vela]

[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 > 0Why would you think it does?[/strike]

 

[flyingpig]

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

That's why I introduce a new inequality?

 

[vela]

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 > 0The 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

 

[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?If I had done

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)

 

[vela]

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?

Not really. So why are you doing it?

If I had done

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)

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.

 

[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.

 

[BruceW]

Because Flyingpig made this rearrangement:
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.

 

[vela]

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.

 

[SammyS]

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)&lt;0 implies that x < 1 .

3. If 0 < a < 1, then \log_{\,a}(x)&lt;0 implies that x > 1 .
​

 

[flyingpig]

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

 

[Mark44]

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)&lt;0 implies that x < 1 .

3. If 0 < a < 1, then \log_{\,a}(x)&lt;0 implies that x > 1 .
​

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

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.