Use the graph of y=2/x^2 - x below to solve the equation 4x^3-10x^2+2=0.

So, [imath]2^x*2^2=2^3*5[/imath]? We still can't solve for [imath]x[/imath].
We're not trying to solve for x algebraically but to use the graph of [imath]2^x[/imath]. Think again.

Side note: It's impractical for an instructor to show you examples for all of the variants of a concept. It's your responsibility to apply what you learned already and build upon it.
 
What is the concept for this, and what are we trying to achieve?
If you have to ask that question, it means you haven't fully understood what we've been doing. You have a record of 4 pages of posts and walked through 3 different problems. The answers are embedded in there and repeated many times. Can you look at them again, and go through the steps? Try to understand why you did what we did, and just not blindly solve equations.

As per the guideline, share your thoughts on the answer to that question, after that we can correct you if needed.
 
No examples have been provided by the instructor, and I haven’t learnt transformation of graphs yet.
You said at one point that you are learning about:
Finding solutions graphically
That means you've been taught something about the sort of problems you are asking for help with.

I want to see what is being taught about that, which I would think would include an example. There probably isn't an example just like the current problem, but there must be something you can show to illustrate what sort of approaches they are taking. If I knew that, I would be in a better position to relate this problem to what you have learned.
 
If you have to ask that question, it means you haven't fully understood what we've been doing. You have a record of 4 pages of posts and walked through 3 different problems. The answers are embedded in there and repeated many times. Can you look at them again, and go through the steps? Try to understand why you did what we did, and just not blindly solve equations.

As per the guideline, share your thoughts on the answer to that question, after that we can correct you if needed.
@BigBeachBanana according to @Dr.Peterson, this problem is a little bit different and won’t work if I use the concept I HAVE LEARNED through the other examples.

You said at one point that you are learning about:

That means you've been taught something about the sort of problems you are asking for help with.

I want to see what is being taught about that, which I would think would include an example. There probably isn't an example just like the current problem, but there must be something you can show to illustrate what sort of approaches they are taking. If I knew that, I would be in a better position to relate this problem to what you have learned.
The textbook I have tells me that I have manipulate the equation until one side matches with the other one. For some reason, the method Steven and you have provided is more thorough for the other problems; I feel more confident using this.

Since y = 2^x and 2^x = 10, I need to intersect y = 2^x with y = 10.

@Dr.Peterson could you show all the methods to solving this exponential function? Transformation of graphs is in my syllabus and will learn it soon.
 
Last edited:
The textbook I have tells me that I have manipulate the equation until one side matches with the other one. For some reason, the method Steven and you have provided is more thorough for the other problems; I feel more confident using this.
It's not clear what that means, exactly. Can you show us exactly what it says?
Since y = 2^x and 2^x = 10, I need to intersect y = 2^x with y = 10.
Yes, that is the simplest approach I can see. What is your answer?
@Dr.Peterson could you show all the methods to solving this exponential function? Transformation of graphs is in my syllabus and will learn it soon.
I've shown you two methods, I believe (left for you to carry out).

Another is to use the graph to find x+2, and then solve that for x.
@BigBeachBanana according to @Dr.Peterson, this problem is a little bit different and won’t work if I use the concept I HAVE LEARNED through the other examples.
Presumably the author thinks that the (general) concept that has been taught should be enough. We want to help you learn to apply such general concepts to novel situations, which is what genuine learning has to include. This is why we have asked you to try. And you have, I think, succeeded.
 
Another is to use the graph to find x+2, and then solve that for x.
Could you show the method of transformation of graphs?

It's not clear what that means, exactly. Can you show us exactly what it says?
I can, but I don't understand the reasoning behind it as I do with the method Steven and you propounded. Here it is:

Yes, that is the simplest approach I can see. What is your answer?
After intersecting, (3.3, 10). So, x ≈ 3.3. Which one is the answer?
 

Attachments

  • Screenshot 2022-08-15 at 7.14.39 PM.png
    Screenshot 2022-08-15 at 7.14.39 PM.png
    199.8 KB · Views: 3
Last edited:
After intersecting, (3.3, 10). So, x ≈ 3.3. Which one is the answer?
We're talking about this problem, right? (Long threads and page boundaries tend to hide things.)
@Dr.Peterson How would I solve 2^(x+2) = 40 if I am given a graph of y = 2^x?
So, [imath]2^x=40/2^2[/imath], which is [imath]2^x=10[/imath].
Yes, the answer is (approximately) x = 3.3. Checking it, we get 2^(3.3+2) = 2^5.3 = 39.397 ≈ 40.

When you ask "which one is the answer?", do you mean, should you say the value of x, or the pair? The solution to an equation in one variable is the value of that variable.

I can [show exactly what it says], but I don't understand the reasoning behind it as I do with the method Steven and you propounded. Here it is:
Clearly this part of one example isn't all the book says about the topic (I was hoping for a general explanation of the procedure); but let's talk about it:

1660572422558.png

This approach does require a lot of insight, and would be much easier with more experience than you have. I think I agree with you that it is hard to generalize; in particular, its applicability depends on the "other side" turning out to be linear so that it can be easily graphed. For the first couple problems you asked about, I do like our way (where we expect to intersect with a linear function and make that happen intentionally) more than theirs (playing around with the equation hoping you get something useful).

But the latest problem requires an entirely different kind of expectation, which is why it would have been very helpful if you had done as we asked earlier, rather than denying that there were any examples at all.

But in fact the general concept being taught here, of manipulating the equation so that it involves the formula for the graph, is directly applicable to the problem we are currently dealing with. And that is what you've done.
 
@Dr.Peterson How would I solve [imath]2*0.5^x+2x-3.5=0[/imath] if I were a given a graph of [imath]y=2*0.5^x-1[/imath]? To be honest, I am actually confused on the standard form of an exponential function. I can't use the laws of indices here.
 
@Dr.Peterson How would I solve [imath]2*0.5^x+2x-3.5=0[/imath] if I were a given a graph of [imath]y=2*0.5^x-1[/imath]? To be honest, I am actually confused on the standard form of an exponential function. I can't use the laws of indices here.
This is an easy one. Do it like the example you showed from the book, or like the first one we did. Don't worry about the exponential function; just compare what you have, [imath]y=2(0.5)^x-1[/imath] to what you want, [imath]2(0.5)^2+2x-3.5[/imath]. Rearrange the latter to have the former on one side.

By the way, as I've hinted, this thread is getting too long, and you shouldn't ask yet another question here. If you have more, ask them separately but include a link to this thread and show your work, so we can remember what ideas you are working with. The only reason to keep asking here is the continuity of ideas, and that's getting tiring.
 
@Dr.Peterson So, [imath]ax+b=2*0.5^x-1[/imath]. [imath](ax+b)/0.5^x=2/0.5^x-1/0.5^x[/imath]
No, no, no. That doesn't help at all.

Just take [imath]2(0.5)^x+2x-3.5=0[/imath] and rearrange it very simply so that it looks like [imath]2(0.5)^x-1 = \text{[something]}[/imath]. This is not hard.

(I see I had a typo before.)

If you want to use my method from before, you want to rearrange [imath]2(0.5)^x-1 = ax+b[/imath] to look like [imath]2(0.5)^x+2x-3.5=0[/imath] and equate coefficients so that they will be the same. But I don't recommend this way, which is not what you are being taught.
 
@Dr.Peterson I know the answer (y=-2x+2.5) using the method of the book. I want to solve it using your method; give me a hint.
Please show your WORK, by whatever method. It may well be exactly what I chose to do for this one.

There is no "my method" in general. The method I suggested for the first problem was ad-hoc, and not something I have even seen before, necessarily. If you mean that, then I already gave you a hint; and another hint was my not recommending that you do it. One way or another, what you need to do is to stop asking, and DO.

But since you choose to be so lazy distrustful of yourself, what I meant by "rearrange [imath]2(0.5)^x-1 = ax+b[/imath] to look like [imath]2(0.5)^x+2x-3.5=0[/imath]" was to change the former to [imath]2(0.5)^x-ax-b-1=0[/imath]. What do a and b have to be in order for those to be identical?
 
I am looking for a way to solve these types of questions. We can't use the first method we learnt to solve exponential functions, and we can't use the method of the book. Is there only one method to stick to?
 
I am looking for a way to solve these types of questions. We can't use the first method we learnt to solve exponential functions, and we can't use the method of the book. Is there only one method to stick to?
Please show your attempts to use each method on the problem you say you can't. Too much has been said in this thread to be sure which problems and which methods you are talking about.

But the only method that always works is THINKING. Every problem may be a little different; if you try to memorize one universal procedure and do the exact same thing with every problem, you will fail on some. That's been one of our points.

My impression is that this is also what the book is trying to do, by not giving you a fully explained procedure, but setting you free to "manipulate" (which is another word for "play"), and learn as you do so. This is how problem solving is taught (though some general guidelines can also be given -- see Polya).

I'll tell you how I came up with my own suggestion for the first problem in the thread: I just visualized what it would mean to use the graph to solve a problem, and figured that the most you could be expected to do was to draw a line intersecting the curve. What line? We have to figure it out. I chose to define the line in terms of two parameters, and see what parameters work. What they did in the one example you've shown us was different; rather than look ahead, they just played with the equation trying to find the formula for the graph hidden in it somewhere.

Another problem may require a different particular way of thinking, but the overall idea is the same: Try things and see what happens. This is why I have been trying so hard to make you do that. You don't learn to play a game by just watching others do it, or listening to what they say.
 
Top