# Reviewing my logarithms, need help solving this inequality

1. Dec 26, 2012

### Astrum

1. The problem statement, all variables and given/known data
$\frac{2^{x+1}-3}{2^{x}-4}\leq1$

2. Relevant equations

3. The attempt at a solution
When I go through it, I keep getting $2^{x}\leq-1$

I don't think the answer is suppose to be complex.... let me show you my work in a file.

It's extremely embarrassing that I don't remember this. Makes me feel pretty stupid....

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2. Dec 26, 2012

### Fightfish

The first step is problematic, because 2^x - 4 could be negative. Multiplying both sides by this factor would then result in the inequality flipping. You should be multiplying both sides by (2^x - 4)^2 instead, which will be a positive quantity.

3. Dec 26, 2012

### Astrum

I'm not sure that I follow. I plugged it into wolfram, and it comes up with a complex solution. If the make it less than, it is real.

Even if the sign flipped, it still can't work out to be real.

4. Dec 26, 2012

### Ray Vickson

Just set $z = 2^x$ and find where $(2z - 3)/(z-4) \leq 1.$ If you can satisfy that with some $z > 0$ you will get a real $x = \log_{2}(z) = \log(z) / \log(2).$

5. Dec 26, 2012

### Fightfish

To remove the fractional term in
$$\frac{2^{x+1}-3}{2^{x}-4}\leq1$$
You must multiply both sides by the denominator squared and not just the denominator because it need not be positive. So, we get
$$(2^{x+1}-3)(2^{x}-4)\leq (2^{x}-4)^2$$
Expand out the terms, collect them on one side, and refactorise.

6. Dec 26, 2012

### haruspex

Isn't it simpler just to treat the two cases separately? The positive sign led to a contradiction, so we can discard that.
No, I think that works. Try it again. If still stuck, please post your working for that case.

7. Dec 26, 2012

### HallsofIvy

Staff Emeritus
Alternatively, do it as two separate problems.
1) if $2^x- 4\ge 0$, multiplying both sides by it gives
$2^{x+1}- 3\le 2^x- 4$
Of course, $2^{x+ 1}= 2(2^x)$ so that inequality is the same as $2(2^x)- 3\le 2^x- 4$. Subtracting $2^x$ from each side and adding 3 to both sides we get $2^x\le -1$ which is, as you say, impossible for x a real number.

2) If $2^x- 4< 0$, multiplying both sides of it changes the direction of the inequality: $2^{x+1}- 3\ge 2^x- 4$ which is the same as $2(2^x)- 3\ge 2^x- 4$. Now, doing the same as before gives $2^x\ge -1$ which is true for all x. Since this was under the condition that $2^x- 4< 0$, the inequality is true for all x satisfying $2^x< 4$ which is the same as saying that x< 2.

8. Dec 26, 2012

### SammyS

Staff Emeritus
You are correct that (2x - 4) is positive for some values of x and negative for others. If Astrum multiplies both sides of the inequality by (2x - 4), then he/she should look at 2 cases separately:
1) 2x > 4, i.e. x>2 .

2) 2x < 4, i.e. x<2 .​

Alternatively, your suggestion of multiplying by (2x - 4)2 eliminates the problem of looking at two cases separately.

Another method, one that's usually covered in College Algebra courses, is as follows:

Subtract 1 from both sides, then use a common denominator to get a single rational expression on the left. That results in a rational expression being compared to zero. For the problem at hand, this resulting rational expression is fairly simple because 2x+1 = 2∙2x .

9. Dec 26, 2012

### Astrum

Ah, Alright. I see. The $\frac{1}{2^{x}-4}$ was where I messed up.

It's been a while since I've seen these problems, I guess some memory lapse is to be expected.

10. Dec 27, 2012

### SammyS

Staff Emeritus
What did you get for a solution ?

11. Dec 27, 2012

### Astrum

x<2

the original problem was wrong, if it meant "equal to" also, because it would lead to a complex answer.

12. Dec 27, 2012

### haruspex

That doesn't make the problem wrong. It asked you to find those x that satisfied the inequality, and you found them.