Reviewing my logarithms, solving this inequality

[Astrum]

Homework Statement

\frac{2^{x+1}-3}{2^{x}-4}\leq1

Homework Equations

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...

 

[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.

 

[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.

 

[Ray Vickson]

Homework Statement

\frac{2^{x+1}-3}{2^{x}-4}\leq1

Homework Equations

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...

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 $x = \log_{2}(z) = \log(z) / \log(2).$

 

[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
instead of whatever you wrote for your second step.
Expand out the terms, collect them on one side, and refactorise.

 

[haruspex]

You must multiply both sides by the denominator squared

Isn't it simpler just to treat the two cases separately? The positive sign led to a contradiction, so we can discard that.

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

No, I think that works. Try it again. If still stuck, please post your working for that case.

 

[HallsofIvy]

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&lt; 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&lt; 0, the inequality is true for all x satisfying 2^x&lt; 4 which is the same as saying that x< 2.

 

[SammyS]

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.

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

1) 2^x > 4, i.e. x>2 .

2) 2^x < 4, i.e. x<2 .​

Alternatively, your suggestion of multiplying by (2^x - 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 2^x+1 = 2∙2^x .

 

[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.

 

[SammyS]

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.

What did you get for a solution ?

 

[Astrum]

What did you get for a solution ?

x<2

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

 

[haruspex]

x<2

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

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