1. ## Junior question

I am not sure if this is the right forum, perhaps it should be in the basic algebra forum but it is the sort of question that involves concepts rather then techniques.

To find a solution set for:

(1): ln(2x+4) = x^2

must one use a graphing method?

Either the intersection of y1 = ln(2x+4) and y2 = (x^2),

or

the intersection of y = ln(2x+4) -x^2 with the x-axis (the zeros of the function)?

If one must use a graphing technique, why is that, what is missing or needed, some new identity?

I note too that in applied mathematics there is such a thing as getting a rough check of your work using "dimensional analysis" but in mathematics, dimensions are not a factor. I can see that this is the case because all expressions made over, say the real numbers, must ultimately reduce to a real number therefore you are always dealing with a correspondence between pure numbers, attached "dimensions" then a separate issue. Still, in this case it does seem that the dividing line between a log equation that is solvable, at least by the basic algebra techniques that I know of, and one that is not, is delineated by mixed dimensions as it were, namely terms that represent exponents (i.e. ln(2x+4) ), with those that do not (i.e. x^2).

In addition to the above question I suppose what I am also sort of grasping for is how one can look at variations of the above equation and know that a graphing solution is the only way to go.

2. ## Re: Junior question

I did move this to the Algebra forum.

Yes, unit analysis in Mathematics can get pretty odd. However Physics can get messy, too. For example, consider the torque equations: 1) [tex]\tau = I \alpha[/math] and 2) $\displaystyle \tau = \vec{r} \times \vec{F}$. 1) has the unit N m rad and 2) has N m. Now radians has the unit 1 rad = 1 m/m, which is unitless but torque is clearly a rotational quantity and thus should contain angular information so a N m rad should not be the same as N m. As well, the unit for energy is 1 J = 1 N m, but torque is clearly not an energy. Basically it's a mess.

As to your equation, yes you will have to use a graphing method or some other numerical technique.

-Dan

3. ## Re: Junior question

Originally Posted by Ray1234
I am not sure if this is the right forum, perhaps it should be in the basic algebra forum but it is the sort of question that involves concepts rather then techniques.

To find a solution set for:

(1): +/1 ln(2x+4) = x^2

must one use a graphing method?

Either the intersection of y1 = ln(2x+4) and y2 = (x^2),

or

the intersection of y = ln(2x+4) -x^2 with the x-axis (the zeros of the function)?

If one must use a graphing technique, why is that, what is missing or needed, some new identity?

I note too that in applied mathematics there is such a thing as getting a rough check of your work using "dimensional analysis" but in mathematics, dimensions are not a factor. I can see that this is the case because all expressions made over, say the real numbers, must ultimately reduce to a real number therefore you are always dealing with a correspondence between pure numbers, attached "dimensions" then a separate issue. Still, in this case it does seem that the dividing line between a log equation that is solvable, at least by the basic algebra techniques that I know of, and one that is not, is delineated by mixed dimensions as it were, namely terms that represent exponents (i.e. ln(2x+4) ), with those that do not (i.e. x^2).

In addition to the above question I suppose what I am also sort of grasping for is how one can look at variations of the above equation and know that a graphing solution is the only way to go.
write your iteration as x = sqrt(ln(2*x + 4)) It took a few iterations to converge to 1.3827

4. ## Re: Junior question

Either way works ... I prefer subtracting the two functions and looking for the zeros.

5. ## Re: Junior question

Originally Posted by votan
write your iteration as x = sqrt(ln(2*x + 4)) It took a few iterations to converge to 1.3827
Okay. Just how did you come up with the iteration formula? That's always been a mystery to me.

-Dan

6. ## Re: Junior question

Hmmm. My title is probably misleading. Sorry.

Thanks for the iteration formula but I had originally put this question in the philosophy form. I am more interested in the nature of the problem rather then a solution, which I do take to be achievable by numerical means.

One mostly studies techniques for solving statements that can be legitimately written. In early training a student (me in particular) assumes that if it is written correctly a statement can be solved. That is, that a statement can always be cooked up from/for a solution set. That turns out to be incredibly naive and idealistic. I am just wondering, if on its face, the above equation can be recognized as solvable only by numerical means and if so what is that feature.

A secondary question was prompted by curiosity. If a mathematician were to attempt to find a method of solving such an equation analytically what would they be looking for?

7. ## Re: Junior question

Dan, here it is

x^2 = ln(2x + 4)
sqrt(x^2) = x = +/- sqrt(ln(2x + 4))
use the positive sqrt first
x = sqrt(ln(2*1 + 4 )) = 1.3386
use this value of x back in the log function
x = sqrt(ln(2*1.3386 + 4 )) = 1.3779
Again, again feed this value back into the log function
x = sqrt(ln(2*1.3779 + 4 )) = 1.3822
iterate
x = sqrt(ln(2*1.3822 + 4)) = 1.3826
again
x = sqrt(ln(2*1.3826)) = 1.3827
x = sqrt(ln(2*1.3826 + 4 )) = 1.33827
Convergence confirmed. This is the positive root.

Do the same with – sqrt to obtain the negative root

Now who told me to do it this way? Nobody. I did not search it and I cannot remember of anyone telling me this iteration. It works with me for almost all transcendental equations. Sometimes it diverges, rewrite it in a reverse order it converges. It was a byproduct from my early research on finding the real zeros of ill conditioned polynomial. At that time I used my calculator hp 15C, very powerful polish logic.

8. ## Re: Junior question

As I think further I see that if I apply both sides of ln(2x+4)=x^2 to e I get, 2x+4 = e^(x^2) which is not an exponential equation since the variable on the left is not the argument of "e" or any other base. Is there a name for this type of equation?

The point is then, that if you don't have an exponential equation you cannot use a log table and log properties to solve it.

I am thinking that if you were to try and invent a technique for solving this sort of problem what you would be trying to do is to find some function other then the log function, say beep(x), that you could apply to each side of the equation that would allow you to consolidate it to a single function equal to a constant, and use beep-1(expression) to return you to "expression", you would also need a beep function table.

I am just thinking aloud and expressing something that might be obvious to most (or obviously wrong) but that I have never seen clearly expressed, nor, personally realized before.

9. ## Re: Junior question

here's a link to some "light" reading that may interest you regarding this topic ...

http://www1.american.edu/cas/mathsta...files/glog.pdf

10. ## Re: Junior question

Yes, this is excellent. A "light" read shows me the development paradigm while a deep read is proving to be perfect "treadmill reading". It is amazing how quickly time on the treadmill passes when you can space out. This is what I was looking for plus some, thanks.