Logarithm problem confused

[Loki123]

Here is the problem and two ways I attempted to solve it. I didn't complete the second one because my question is can I do it the first way, or do I have to do it the second way? Did I solve it by accident by first way (entire first pic) or do I have to do it the second way (entire second pic)?


I don't have the original source of the problem.

 

[Dr.Peterson]

Here is the problem and two ways I attempted to solve it. I didn't complete the second one because my question is can I do it the first way, or do I have to do it the second way? Did I solve it by accident by first way (entire first pic) or do I have to do it the second way (entire second pic)?

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I don't have the original source of the problem.

Is there a reason you think only the second way is legal? It's normal to have multiple valid methods.

The only thing I see wrong with the first is that, by not using any words, you have not made all the reasoning clear. I think you are relying on the monotonicity of the log in two different cases, which you need to state explicitly.

 

[JeffM]

There used to be a helper here who would sensibly say, “Correct answers do not care how you find them.”

It is true that x must exceed 0 because the argument of any logarithm must exceed 0. But there are two restrictions on the base of a logarithm, namely it must exceed 0 and must not equal 1.

[math]x - 2 > 0 \implies x > 2.\\ x - 2 \ne 1 \implies x \ne 3.[/math]
There is nothing wrong with the general line of attack in method 1, but

[math]x = e - 2 \implies x + 2 = e.\\ ln(4) > 1 = ln(e) = ln(x + 2).[/math]
Perhaps you should rethink your answer

 

[Loki123]

Is there a reason you think only the second way is legal? It's normal to have multiple valid methods.

The only thing I see wrong with the first is that, by not using any words, you have not made all the reasoning clear. I think you are relying on the monotonicity of the log in two different cases, which you need to state explicitly.

I feel like I shouldn't just dismiss x-2 like I did.

 

[JeffM]

I feel like I shouldn't just dismiss x-2 like I did.

As Dr. Peterson said, there is another aspect to this question that needs attention,

[math]a > 1 \text { and } \log_a(p) > log_a(q) \implies p > q.[/math]
So once you know that x > 2, all that work in method 1 and method 2 is unnecessary.

[math]\log_{x-2}(4) \ge \log_{x-2}(x) \implies 4 \ge x.[/math]
(I did not consider bases between 0 and 1. I always have to think those through from first principles, and they are irrelevant for this problem.)

 

[Dr.Peterson]

I feel like I shouldn't just dismiss x-2 like I did.

If you just "dismissed" the base without thinking, then your thinking must be wrong. But it looks to me as if you did pay attention to it, in forming your two cases, namely [imath]x-2>1\implies x>3[/imath], and [imath]0<x-2<1\implies 2<x<3[/imath].

Perhaps you need to tell us the details of your thinking, so we don't have to guess what you had in mind, as I did.

 

[JeffM]

If you just "dismissed" the base without thinking, then your thinking must be wrong. But it looks to me as if you did pay attention to it, in forming your two cases, namely [imath]x-2>1\implies x>3[/imath], and [imath]0<x-2<1\implies 2<x<3[/imath].

Perhaps you need to tell us the details of your thinking, so we don't have to guess what you had in mind, as I did.

But of course the stipulation is that [imath]x - 2 > 0, \text { not } x - 2 > 1.[/imath]

What he remembered is that the base must not equal 1. Look at my example where x = e - 2.

 

[Loki123]

But of course the stipulation is that [imath]x - 2 > 0, \text { not } x - 2 > 1.[/imath]

What he remembered is that the base must not equal 1. Look at my example where x = e - 2.

From what I learned, if it's >1, sign stays the same; it's between 1 and 0, sign switches.

 

[Dr.Peterson]

From what I learned, if it's >1, sign stays the same; it's between 1 and 0, sign switches.

I think what you mean is that if the base is greater than one, the log is a monotonically increasing function, and if it is between 0 and 1, the log is a decreasing function. Yes, that is the fact I assumed you were using. Please write out your reasoning using this fact, so that your solution will communicate clearly.

But of course the stipulation is that [imath]x - 2 > 0, \text { not } x - 2 > 1.[/imath]

What he remembered is that the base must not equal 1. Look at my example where x = e - 2.

I think you're misinterpreting what we are both referring to.

 

[JeffM]

I see I managed to mess up my posts # 3 and # 5 big time.

[math]\text {Find domain of } x \text { given } \log_{x-2}(4) \ge \log_{x-2}(x).[/math]
The proposed solution did NOT address two cases. What was specifically asserted was that [imath]x - 2 > 1.[/imath]

But nothing in the problem that was presented requires that condition.

[math]x - 2 > 0 \implies x > 2.\\ x - 2 \ne 1 \implies x \ne 3.[/math]
[math]\text {If } 1 < a \text { and } p > 0 < q, \text { then } \log_a(p) > \log_a(q) \iff p > q.\\ \therefore x > 3 \text { and } \log_{x-2}(4) \ge \log_{x-2}(x) \implies x - 2 > 1 \implies 4 \ge x.[/math]
Quite straight forward. So far so good. But the second potential case of [imath]2 < x < 3 [/imath] must be addressed. At this point, my brain went off the rails, probably because I hate thinking about bases less than 1.

Is it impossible to meet the condition [imath]\log_{x-2}(4) \ge \log_{x-2}(x)[/imath] if [imath]2 < x < 3[/imath]?

That is not intuitively obvious to me. And it certainly was not addressed in post 1. Of course I did not address it in posts 3 and 5.

[math]a = x - 2 \text { and } 2 < x < 3 \implies 0 < a < 1.\\ \text {Let } b = 1/a \implies b > 1 \text { and } \log_b(a) = - 1.\\ \therefore \log_a(x) = \dfrac{\log_b(x)}{\log_b(a)} = - log_{b}(x)\\ \text {and } \log_a(4) = \dfrac{\log_b(4)}{\log_b(a)} = - \log_{b}(4).\\ \therefore \log_a(4) \ge \log_a(x) \implies - \log_b(4) \ge - \log_b(x) \implies\\ \log_b(4) \le \log_b(x) \implies 4 < x \ \because \ b > 1.\\ \text {But that is inconsistent with the hypothesis that } 2 < x < 3.\\ \therefore \text { It is impossible that } 2 < x < 3 \text { and } \log_{x-2}(4) \ge \log_{x-2}(x).[/math]
The proposed solution is correct, but the logic, whether method 1 or 2, was incomplete.