Logarithm problem how (ref: https://www.freemathhelp.com/forum/threads/logarithm-problem-confused.133645/)

[Loki123]

The first line is the problem, the second line is how the first line is supposed to be transformed so we can solve it easier. I have no idea how to do this. How do you I get the second line? I know how to solve it, I am not looking for alternative ways. I want to understand this way.

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

 

[Deleted member 4993]

The first line is the problem, the second line is how the first line is supposed to be transformed so we can solve it easier. I have no idea how to do this. How do you I get the second line? I know how to solve it, I am not looking for alternative ways. I want to understand this way.
View attachment 31620
I don't have the original source of the problem.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Loki123]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

I am stuck at the beginning. I have no idea what to do next. There is literally nothing else I could post, sorry.

 

[AvgStudent]

The first line is the problem, the second line is how the first line is supposed to be transformed so we can solve it easier. I have no idea how to do this. How do you I get the second line? I know how to solve it, I am not looking for alternative ways. I want to understand this way.
View attachment 31620
I don't have the original source of the problem.

I can't speak as to why how to go from line 1 to 2, but I don't think it's necessary to do that to solve for the domain of x.

1. [imath]\log_{2}(x-2) \implies x>2[/imath]
2. The denominator cannot be 0 and it is 0 when [imath]\log_{2}(x-2)=0 \implies 2^0=1=x-2 \implies x=3[/imath], so [imath]x\neq 3[/imath]
3. Overall, the fraction is greater or equal to 0 when [math]\dfrac{\log_{2}(x)-2}{\log_{2}(x-2)}\implies \log_{2}(x)-2 \ge 0 \implies x \ge 4[/math]

Finally, you just need to put the pieces together

 

[Loki123]

I can't speak as to why how to go from line 1 to 2, but I don't think it's necessary to do that to solve for the domain of x.

1. [imath]\log_{2}(x-2) \implies x>2[/imath]
2. The denominator cannot be 0 and it is 0 when [imath]\log_{2}(x-2)=0 \implies 2^0=1=x-2 \implies x=3[/imath], so [imath]x\neq 3[/imath]
3. Overall, the fraction is greater or equal to 0 when [math]\dfrac{\log_{2}(x)-2}{\log_{2}(x-2)}\implies \log_{2}(x)-2 \ge 0 \implies x \ge 4[/math]

Finally, you just need to put the pieces together

Thank you, but I did imply in my post that I know how to solve it, I just want to understand this way.

 

[AvgStudent]

Thank you, but I did imply in my post that I know how to solve it, I just want to understand this way.

My apologies. I think they did it almost similar to what I did.
For the denominator, they equated equal to 0 because we can't divide by 0.
[imath]\log_{2}(x-2)=0 \implies 2^0=1=x-2 \implies x-3=0[/imath]
Similarly, they equated the numerator
[imath]\log_{2}x-2\ge 0 \implies\log_{2}x\ge 2 \implies 2^{\log_{2}x} \ge 2^2 \implies x\ge 4 \implies x-4\ge 0[/imath].
Together, [imath]\dfrac{x-4}{x-3}\ge 0[/imath]. However, I think this is a bad way to do the problem because you can easily overlook that [imath]x>2[/imath] is also a constraint from the original problem.

 

[Dr.Peterson]

The first line is the problem, the second line is how the first line is supposed to be transformed so we can solve it easier. I have no idea how to do this. How do you I get the second line? I know how to solve it, I am not looking for alternative ways. I want to understand this way.
View attachment 31620
I don't have the original source of the problem.

Are you saying that someone reputable started with this step, with no explanation? That's not very good communication!

It does look like @AvgStudent's idea is probably more or less the thinking that lies behind this, but it needs to be stated more clearly.

The initial inequality will be true when the numerator and denominator have the same sign, or when the numerator is zero.

Since [imath]\log_2x-2[/imath] is an increasing function that is zero when [imath]x=4[/imath], the sign of [imath]\log_2x-2[/imath] is the same as the sign of [imath]x-4[/imath].

Since [imath]\log_2(x-2)[/imath] is an increasing function that is zero when [imath]x=3[/imath], the sign of [imath]\log_2(x-2)[/imath] is the same as the sign of [imath]x-3[/imath].

We also have the condition that [imath]x>2[/imath] so that both logs exist, which they have failed to mention.

So it is true that (apart from [imath]x>2[/imath]), the second line is equivalent to the first; but I see no reason to actually write that second line, when the next step (as I am guessing it should be) will be to use the facts I stated, rather than the second line which is based on those facts.

But it would have been far better if you had shown us the entire solution, not just these two lines, because then we could see where they are heading and hopefully more quickly understand why they are saying this. Please don't do this to us in the future. Show the entire problem and the entire work, even when you are only asking (you think) about one piece. Context often makes a big difference.

 

[Loki123]

Are you saying that someone reputable started with this step, with no explanation? That's not very good communication!

It does look like @AvgStudent's idea is probably more or less the thinking that lies behind this, but it needs to be stated more clearly.

The initial inequality will be true when the numerator and denominator have the same sign, or when the numerator is zero.

Since [imath]\log_2x-2[/imath] is an increasing function that is zero when [imath]x=4[/imath], the sign of [imath]\log_2x-2[/imath] is the same as the sign of [imath]x-4[/imath].

Since [imath]\log_2(x-2)[/imath] is an increasing function that is zero when [imath]x=3[/imath], the sign of [imath]\log_2(x-2)[/imath] is the same as the sign of [imath]x-3[/imath].

We also have the condition that [imath]x>2[/imath] so that both logs exist, which they have failed to mention.

So it is true that (apart from [imath]x>2[/imath]), the second line is equivalent to the first; but I see no reason to actually write that second line, when the next step (as I am guessing it should be) will be to use the facts I stated, rather than the second line which is based on those facts.

But it would have been far better if you had shown us the entire solution, not just these two lines, because then we could see where they are heading and hopefully more quickly understand why they are saying this. Please don't do this to us in the future. Show the entire problem and the entire work, even when you are only asking (you think) about one piece. Context often makes a big difference.

After what I have posted all that follows is:

X>0
X>2
X can't be 3
X is (3,4]

 

[Dr.Peterson]

After what I have posted all that follows is:

X>0
X>2
X can't be 3
X is (3,4]

Thanks. It really isn't very well explained, is it? I hope this isn't from a teacher.

I was right, I think, that the supposedly transformed line is not really used, but rather the reasoning that led to it. The first two lines here state the conditions on the domains of the two logs and on the denominator, and then the conclusion, which must be based on the inequalities used to write the rational equation, not on that equation itself.

Here is the whole answer as I might put it:

The initial inequality will be true when the numerator and denominator have the same sign, or when the numerator is zero.​

​

Since [imath]\log_2x-2[/imath] is an increasing function that is zero when [imath]x=4[/imath], the sign of [imath]\log_2x-2[/imath] is the same as the sign of [imath]x-4[/imath].​

​

Since [imath]\log_2(x-2)[/imath] is an increasing function that is zero when [imath]x=3[/imath], the sign of [imath]\log_2(x-2)[/imath] is the same as the sign of [imath]x-3[/imath].​

​

Therefore either [imath]x>4[/imath] and [imath]x>3[/imath], so that [imath]x>4[/imath]; or [imath]x<4[/imath] and [imath]x<3[/imath], so that [imath]x<3[/imath].​

​

In order for the logs to be defined, we also need [imath]x>0[/imath] and [imath]x>2[/imath].​

​

The value will be 0 when the numerator is zero, so we can include [imath]x=4[/imath], but not [imath]x=3[/imath].​

​

Putting these inequalities together, [imath]x\ge 4[/imath] or [imath]x<3[/imath], and [imath]x>2[/imath], resulting in the interval [imath](2,3)\cup[4,\infty)[/imath].​

​

I never followed through to check the solution, so now I need to ask whether you copied the problem incorrectly. Did they really solve it with [imath]\le[/imath] instead of [imath]\ge[/imath]? Here is the graph:

​

​

 

[Loki123]

Thanks. It really isn't very well explained, is it? I hope this isn't from a teacher.

I was right, I think, that the supposedly transformed line is not really used, but rather the reasoning that led to it. The first two lines here state the conditions on the domains of the two logs and on the denominator, and then the conclusion, which must be based on the inequalities used to write the rational equation, not on that equation itself.

Here is the whole answer as I might put it:

The initial inequality will be true when the numerator and denominator have the same sign, or when the numerator is zero.​

​

Since [imath]\log_2x-2[/imath] is an increasing function that is zero when [imath]x=4[/imath], the sign of [imath]\log_2x-2[/imath] is the same as the sign of [imath]x-4[/imath].​

​

Since [imath]\log_2(x-2)[/imath] is an increasing function that is zero when [imath]x=3[/imath], the sign of [imath]\log_2(x-2)[/imath] is the same as the sign of [imath]x-3[/imath].​

​

Therefore either [imath]x>4[/imath] and [imath]x>3[/imath], so that [imath]x>4[/imath]; or [imath]x<4[/imath] and [imath]x<3[/imath], so that [imath]x<3[/imath].​

​

In order for the logs to be defined, we also need [imath]x>0[/imath] and [imath]x>2[/imath].​

​

The value will be 0 when the numerator is zero, so we can include [imath]x=4[/imath], but not [imath]x=3[/imath].​

​

Putting these inequalities together, [imath]x\ge 4[/imath] or [imath]x<3[/imath], and [imath]x>2[/imath], resulting in the interval [imath](2,3)\cup[4,\infty)[/imath].​

​

I never followed through to check the solution, so now I need to ask whether you copied the problem incorrectly. Did they really solve it with [imath]\le[/imath] instead of [imath]\ge[/imath]? Here is the graph:

​

View attachment 31634​

I just checked and yes the sign is on the wrong side. I am very sorry about that. I must have got confused and wrote that from memory. I'll see that it does not happen again.