A Mathematical Emergency.

[Don Blazys]

In this topic, I will be making an extraordinarily "strange claim", almost unbelievable in fact, and it will be up to you, the reader, to decide if the properties of logarithms, strange as they may be, are telling us the truth!

 

So hold on to your hats and buckle your saftey belts, because you are about to embark on the strangest, wildest, and perhaps most wonderfull mathematical ride of your life!

 

Let all variables herein represent non-negative integers. Then, the recently discovered identity:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))

 

shows that it is algebraically impossible to "cross out" the cancelled T's. What does this mean to you? Should we stop teaching students to "cross out" cancelled factors and common factors? I say yes, and will present my reasons for doing so as this thread continues. What do you say?

 

Don.

[Jay-qu]

Have you heard of latex? Try giving that a go and it will make your maths more legible. At the same time you may as well try and give a more mathematical formalism of what it is you are trying to prove/disprove. Let me know how you get on.

[CraigD]

Let all variables herein represent non-negative integers. Then, the recently discovered identity:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))

If the above expression assumes the usual precedence of operations (eg: exponentiation before multiplication), then the given equation is provably false.

 

As Jay notes, math is much more readable using hypography’s available LaTeX package, so I’ll translate Don’s text to that for this very short proof:

[math]\left( \frac{T}{T}\right)a^x \overset{?}{=} T\left( \frac{a}{T}\right) ^{\left( \frac{\left( \frac{x \ln(a)}{\left(ln(T)-1\right)} \right)}{\left( \frac{\ln(a)}{\left(ln(T)-1\right)} \right)} \right)}[/math]

removing identities (ones),

[math]a^x \overset{?}{=} T\left( \frac{a}{T}\right)^x[/math]

expanding and combining common coefficients,

[math]a^x \overset{?}{=} T^{x-1}a^x[/math]

dividing both sides of equation by [math]a^x[/math],

[math]1 \overset{?}{=} T^{x-1}[/math]

we find

[math]1 \not= T^{x-1}[/math]

so,

[math]\left( \frac{T}{T}\right)a^x \not= T\left( \frac{a}{T}\right) ^{\left( \frac{\left( \frac{x \ln(a)}{\left(ln(T)-1\right)} \right)}{\left( \frac{\ln(a)}{\left(ln(T)-1\right)} \right)} \right)}[/math]

except for special cases [math]T=1[/math] and [math]x=1[/math].

… shows that it is algebraically impossible to "cross out" the cancelled T's. What does this mean to you?

Because the first given is false, no further comments are formally meaningful. :(

Should we stop teaching students to "cross out" cancelled factors and common factors? I say yes, and will present my reasons for doing so as this thread continues. What do you say?

I say yes, too, but with some qualification.

 

From several years of tutoring college Math, a combined couple of years of teaching remedial college and GED-track high school Math, and the common parental experience of helping several now adult children (my own and others) through grades K-12, I think there’s a great need to acquaint students with the idea of mathematical formalism – that is, the idea that, as someone put it, “algebra is a collection of exact rules for manipulating marks on paper” – at nearly the earliest age possible, 5 or 6 years.

 

Reliance on phrases like “cross out” and “cancel”, in my experience, usually show a lack of grasp of the fundamentals of formalism, a lack that can handicap students in Math and a large collection of similar disciplines for the rest of their lives. Unfortunately, in the present-day US, and, I suspect, many other states’, schools, formalism is usually introduced only in the last years of high school or later years of college (ages 16 to 21), and then only to students with specializing in Math-intense education paths.

 

IMHO, a strong grasp of formalism needs to be as essential as knowing ones alphabet, and taught as early and as well. My opinion is by no means original or untried, dating back at least to suggestions from turn-of-the-20th-century mathematicians and educators, and having made sporatic widespread appearances in curriculum under monikers such as “new math” for over 50 years, as well as being taught in isolated instances in many schools and by many teachers within and outside of traditional schools. My experience with attempting to promote this approach, however, has taught me that it’s not an easy task. It’s necessary to first win the hearts and minds of professional educators, who are often themselves math-adverse to the extent of near innumeracy. Learned early and well, formalism is, in my experience, an easy concept to understand and apply, but studied late in ones education, very difficult to really grasp and appreciate.

 

PS: Don, you can pick up the essentials of hypography’s implementation of the LaTeX math package and markup code tags by clicking “quote” on this post and inspecting the text. More in depth documentation on LaTeX can be found in many places, including a couple of my favorites, the wikimedia article “Help: Displaying a formula” and this easy-to-use tutorial. For documentation on hypography's supported markup tags like [math], see the BB Code List FAQ page.

 

In math, presentation – pretty handwriting, or its computerized equivalent – can be very effective, and is IMHO worth making a strong effort to learn well in whatever medium you find yourself. :)

[Don Blazys]

To: Jay-qu. Due to circumstances beyond my control, I can't post in LaTex at this time. However, if you keep in mind that operations are to be performed from the intermost parenthesis outward, then you should have no trouble understanding the identity:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)).

 

You know, when I was in high school back in the 60's, we didn't have personal computers, calculators and the like. We used pencil, paper and slide rule to solve our math problems, and although my generation developed much of the technology in use today, we never became overly dependent on it. Just for the fun of it, why don't you give it a go and see if you can verify the above identity without me putting it into LaTex?

 

Don.

[Don Blazys]

To: CraigD. The equation:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))

 

is both true and correct. It is not false!

You made a mistake when you put it into LaTex.

Note that the expression: (ln(T)) is a divisor, not a multiplier.

Please try putting it into LaTex again, but with the correct operations this time.You will then find that it is an identity, and that it is indeed algebraically impossible to "cross out" the cancelled T's.

 

Don.

[Jay-qu]

To: Jay-qu. Due to circumstances beyond my control, I can't post in LaTex at this time.

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)).

 

Just for the fun of it, why don't you give it a go and see if you can verify the above identity without me putting it into LaTex?

 

Don.

 

LateX only helps the output look nice, it doesnt make the identity any easier to solve. Its just nicer to read as craig has shown - and no its not like doing things with pencil and paper is harder, in fact I much rather it for maths. Displaying maths on the computer is much slower and cumbersome than on paper.

[Qfwfq]

And I did not need to work it onto paper, nor reach Craig's post, to see it is not an actual identity, and when I got to Craig's post I noticed some slips which Don himself didn't point out.

 

Don I see no way for the exponent to identically equal 1 independently of x and if there's any way of the logarithms in the exponent compensating the T's in the base, go ahead and show us. You will not increase your audience by harping about pencil and paper in the good ole days instead of making the effort to use LaTeX.

[CraigD]

To: CraigD. The equation:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))

 

is both true and correct. It is not false!

As I trivially prove in post #2, the equation is true only for the special cases of [math]x=1[/math] or [math]T=1[/math]

 

The flaw with the proposed identity is that exponentiation isn’t associative with multiplication, that is [math]abc^d \not= a(cb)^d[/math].

You made a mistake when you put it into LaTex.

I did. :( My apologies – I’ve corrected it.

 

Since the exponent in question still equate to [math]x \cdot 1[/math], however, the difference has no impact on the proof.

Due to circumstances beyond my control, I can't post in LaTex at this time.

No special software is needed to use the LaTeX math package in hypography posts, just the post entry box, or whatever text editor you use to make posts. Once you’ve learned a few basic functions, such as \frac{a}{b} as an equivalent of a/b, you’ll find LaTeX text it’s very similar to the text you’ve already written. For example, the LaTeX for

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))

is

\left( \frac{T}{T}\right)a^x = T \left( \frac{a}{T} \right) ^{\left( \frac{\left( \frac{x \ln(a)}{\left( \ln(T)-1\right)} \right)}{\left( \frac{\ln(a)}{\left( \ln(T)-1\right)} \right)} \right)}

If the parenthesis made unnecessary by the other markups are omitted, it’s an even simpler

\frac{T}{T}a^x = T \left(\frac{a}{T} \right)^{ \frac{\frac{x \ln(a)}{\ln(T)-1}}{\frac{\ln(a)}{\ln(T)-1}} }

You know, when I was in high school back in the 60's, we didn't have personal computers, calculators and the like. We used pencil, paper and slide rule to solve our math problems, and although my generation developed much of the technology in use today, we never became overly dependent on it.

We didn’t have PCs when I was in high school in the ‘70s, either. Though there are some input devices that manage to make a computer fairly equivalent to paper and pencil, I, and nearly everyone I know, still use paper and pencil for most math and graphical sketching. Only when “neatening up” for posting do we transcribe into LaTeX or another computer displayable form.

 

An alternative to using LaTeX or similar math markup schemes is to scan paper documents and upload them as images or thumbnails in your posts. You’ll notice that members such as Turtle (who writes a lot of complicated, pretty things that can’t be rendered in LaTeX) do this quite a bit. For ordinary math expressions, however, most people find it easier to use LaTeX, especially folk who don’t have easy access to a scanner.

[Don Blazys]

To:Qfwfq. The identity:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))

 

is true! It is however, quite counterintuitive, and most people have a very hard time believing it. One professor even went so far as to graph both sides on a supercomputer, then called me the next day to express his utter amazement. So please keep at it. Once you understand it, I guarantee that you will also be amazed. You can find the formal derivation of the above equation on my website (donblazys.com), as well as a lively discussion about it on the Marilyn Vos Savant general discussion forum.

 

Don.

[Don Blazys]

To:CraigD. The identity:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))

 

is true. Please keep in mind that the logarithms can be "crossed out" if and only if x=1. For all other values of x, the logarithmic exponent must remain exactly as it is. In other words, don't confuse the above logarithmic exponent, which is:

 

((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) for

 

x((ln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)).

 

There is also a lot more information about this equation, including its derivation, on my website (donblazys.com).

 

As for putting the above equation in LaTex, I really appreciate all the help that you have given me so far. However, the problem that I am having is this:

While on the "reply to thread" board, after I type in the "commands":

[math] \left(\frac{T}{T}\right)a^x....etc.,

I can't get those "commands" to convert my equation into LaTex.

Any suggestions?

 

Gotta go now, but I will be back on Monday. Have a great weekend!

 

Don.

[Don Blazys]

To: CraigD. P.S. In LaTex, the logarithmic exponent should assume the form:

 

 

xln(a)

--------- -1

ln(T)

----------------------

ln(a)

--------- -1

ln(T)

 

 

 

 

Please use the above to correct your initial post on this thread.

 

Don.

[CraigD]

To: CraigD. P.S. In LaTex, the logarithmic exponent should assume the form…

Mea maxima culpa! :( My apologies for taking so long to get it right. :embarass: All those parentheses got me mixed up – embarrassing for a guy who writes parsers and other parentheses-y things for a living. :embarass: Hopefully, we’ve now gotten past the electronic barrier of writing the equation (which would have taken all of 30 seconds using an old-fashioned chalkboard). Though, to spread the culpa around a bit, Don could have saved me from my folly by learning a bit of LaTeX. It’s not difficult, and offers the pleasure at seeing ugly keystrokes rendered as pretty pictures. :)

 

Here’s the equation correctly rendered:

[math]\frac{T}{T}a^x = T \left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}[/math]

 

A bit more compactly:

[math]a^x = T \left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}[/math]

 

Dividing both sides by [math]T[/math], this can be written

[math]\frac{a^x}{T} = \left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}[/math]

I’m not sure if this is what the Don meant in the original post by “cross out the canceled Ts”. I don’t see anything algebraically abnormal about the equation :shrug:

 

It is a remarkable identity, and looks like fun to prove. It’s pretty easy to demonstrate, running a little MUMPS program (with a standard, 18-digit precision calculator) on my cheap 4-year-old laptop (no supercomputer required :)):

f  r "a=",a," T=",T," x=",x," " s E=x*$zln(a)/$zln(T)-1/($zln(a)/$zln(T)-1) w a,"^",x,"=",a**x," ",T,"(",a,"/",T,")^(",E,")=",T*(a/T**E),!
a=1 T=2 x=3 1^3=1 2(1/2)^(1)=1
a=2 T=3 x=4 2^4=16 3(2/3)^(-4.128533874054364329)=15.99999999999999945
a=2 T=4 x=3 2^3=8 4(2/4)^(-1)=8
a=4 T=2 x=3 4^3=64 2(4/2)^(5.000000000000000001)=63.9999999999938609
a=9 T=3 x=2 9^2=81 3(9/3)^(2.999999999999999998)=80.99999999999948382
a=16 T=4 x=2 16^2=256 4(16/4)^(3)=256
a=25 T=5 x=2 25^2=625 5(25/5)^(3)=625
a=8 T=2 x=2 8^2=64 2(8/2)^(2.5)=63.99999999999999652
a=.25 T=.5 x=2 .25^2=.0625 .5(.25/.5)^(3.000000000000000001)=.0624999999999969838
a=.5 T=.6 x=.7 .5^.7=.6155722066724512082 .6(.5/.6)^(-.1405352050771790835)=.6155722066724661252
a=.5 T=3 x=2 .5^2=.25 3(.5/3)^(1.386852807234541587)=.2499999999999999888
a=2.718281828459045236 T=7.389056098930650228 x=2 2.718281828459045236^2=7.389056098930650231 7.389056098930650228(2.718281828459045236/7.389056098930650228)^(-.000000000000000002)=7.389056098930649799
a=7.389056098930650228 T=2.718281828459045236 x=3 7.389056098930650228^3=403.4287934927351227 2.718281828459045236(7.389056098930650228/2.718281828459045236)^(5.000000000000000002)=403.4287934927350995

The difference between the two sides of the equation at more than 18 significant digits are typical of rounding error when approximating transcendental functions such as logarithms, which are used not only in the natural log focution ([math]\ln[/math], $zln), but also for non-integer exponentiation.

[CraigD]

Here’s a proof of

[math] a^x \overset?= T \left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}[/math]

 

[math]\log_a b = \frac{\log a}{\log b}[/math], so, for ease of writing, rewrite as

[math]a^x \overset?= T \left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -1}{\log_Ta -1}\right)}[/math]

 

Taking the logarithm base a of both sides gives:

[math]\log_a\left(a^x \right) \overset?= \log_a\left(T \left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -1}{\log_Ta -1}\right)} \right)[/math]

 

[math]\log\left(a^b\right)=b\log a[/math], [math]\log_a a = 1[/math], and [math]\log\left(ab\right)=\log a +\log b[/math] give

[math]x \overset?= \log_aT +\frac{x\log_Ta -1}{\log_Ta -1}\log_a\left(\frac{a}{T}\right)[/math]

 

And

[math]x \overset?= \log_aT +\frac{x\log_Ta -1}{\log_Ta -1} \left(1 -\log_aT \right)[/math]

 

Multiply both sides by [math] \log_Ta -1[/math] giving

[math](\log_Ta -1)x \overset?= (\log_Ta -1)\log_aT +(x\log_Ta -1)(1 -\log_aT)[/math]

 

Expand, giving

[math]x\log_Ta -x \overset?= \log_aT\log_Ta -\log_aT +x\log_Ta -1 -x\log_aT\log_Ta +\log_aT[/math]

 

[math]\log_ab\log_ba = 1[/math], so

[math]x\log_Ta -x \overset?= 1 -\log_aT +x\log_Ta -1 -x +\log_aT[/math]

 

And simplify, giving

[math]-x = -x[/math]

[Don Blazys]

To: CraigD. Thats good. Just exellent! You have more than redeemed yourself! I too apologize and take full responsibility for not being able to write it in LaTex. I work at a high school, and have now resolved to get one of the computer teachers to show me how to do it the first chance I get. As you could probably surmise, I'm quite inexperienced at working with computers, so I greatly appreciate your hard work and patience, and thank you for seeing it through rather than giving up like so many others have done.

 

Since we have now established that the first equation in post #12 is indeed an identity, I can promise that from here on out, that identity will make for a very interesting discussion, mainly because it is relatively unknown and not yet in any book, journal, magazine or encyclopedia. You see, I discovered it only about a decade ago, and it's consequences and ramifications have yet to be fully addressed by the math community. In fact, it's been only in the past few months that the math departments at two major universities and the editors of a major math journal have begun a serious study of this rather remarkable identity, so it is clear that we are all entering into "uncharted mathematical territory", where there is no "general consensus" on some very basic issues.

 

Personally, I find this all very exiting because this stuff is both "elementary", and "cutting edge", and just as the early experiments involving quantum mechanics required "intelligent interpretation", so does this relatively new identity!

 

For instance, since the first equation in post #12 is a proven identity, and since that identity is true for all values of the variables T={2,3,4...}, a={1,2,3...} and x={0,1,2...}, is the term on the right more suitable for representing "common factors" than the term on the left? I say yes, because "common factors" absolutely must be defined as being greater than unity in order for that concept to make any sense whatsoever, and the term on the right clearly precludes (prevents) T=1 when T is to be construed as a "common factor". What do you say?

 

Also, is it now proper to "cross out" the cancelled T's in the term on the left as you did in the second equation in post#12? I have a problem with that because once the T's are "crossed out", the term on the left appears as a^x, which can be construed as (1)a^x=(N/N)a^x where N does not equal T, and while the equation would remain true, it would no longer be an identity, and we would be either ambiguous (in the case of (1)a^x), or inconsistent (in the case of (N/N)a^x) as to the value of the cancelled factor or cancelled common factor. Again, what do you say?

 

There are other, even more serious issues, but let's discuss these two first. Maybe others will join us.

 

P.S. What kind of calculator do you have that is accurate to 18 decimal places, or is such accuracy available only in conjunction with a computer? Are they expensive?

 

Thanks again my mathematical friend!

 

Don.

[CraigD]

Easy questions first…

P.S. What kind of calculator do you have that is accurate to 18 decimal places…

I’m using the built in calculator of an implementation of the MUMPS programming language (specifically, Intersystems Caché) that I usually have a terminal session connected to. If I recall correctly, the language standard requires it give at least 16 digits precision, and most do better.

 

I’ve written my own calculators in MUMPS for when I must be able to, for example, do precise arithmetic with million digit numbers. MUMPS is traditionally a database-focused language, used a lot in medicine and banking, so I’m a bit of an oddball for using it for math and science, although it has interesting and fairly unappreciated, IMHO, potential in these disciplines.

 

MUMPS is a programming languages, so a bit much to install and manage just to have a decent calculator, though very handy if you do, and learn just a bit about using it. You may have notices in some of my posts that you can write pretty useful MUMPS programs with a very small number of keystrokes.

 

There are several applications that include high or arbitrary precision calculators, such as Mathematica and Maple. Apps like these do a lot more than just literal calculations – I find that they make me feel rather stupid, as in a sense, they “know” much more math than I do.

 

Many folk find the ruby language handy for use as a calculator Lots about it, including documentation, tutorials, and a free Windows implementation of it, can be found here. If you can’t or don’t want to install it on your computer, you can use it via a browser here.

... or is such accuracy available only in conjunction with a computer?

Short answer, yes, you need a computer capable of running your calculator app of choice. Though in principle you can run nice calculator apps on the average cellphone these days, in practice this is pretty tricky. Mostly, I use a computer as a calculator. I’m rarely without a computer of a sort, having a PalmOS handheld in my pocket most of the time, though my calculators on it aren’t as good as on a Windows or unix machine. I hope someday soon to have about the same resources on my handheld as on a normal PC - I’m not quite there yet, but may be soon, as there are a growing number of handhelds being brought to market that run general-purpose operating systems, typically linux.

 

On the subject, you might find the hypo thread 3998 interesting.

Are they expensive?

Assuming you have a PC, you can get free copies of Caché or several other implementation of MUMPS (there’s also an open source version). Ruby is also free, via the links above.

 

Mathematica and Maple are a bit pricey, though if you’re with a school, you can usually get a good discount.

 

Now for a harder question…

Also, is it now proper to "cross out" the cancelled T's in the term on the left as you did in the second equation in post#12? I have a problem with that because once the T's are "crossed out", the term on the left appears as a^x, which can be construed as (1)a^x=(N/N)a^x where N does not equal T, and while the equation would remain true, it would no longer be an identity, and we would be either ambiguous (in the case of (1)a^x), or inconsistent (in the case of (N/N)a^x) as to the value of the cancelled factor or cancelled common factor. Again, what do you say?

Algebraically, it’s always proper to replace a term of the form [math]\frac{x}{x}[/math] with 1. When doing so, it’s important to retain any domain constraints of the replaced term (eg: [math]x \not= 0[/math])

 

Usually, one tries to write math canonically, replacing any terms equal to 1, so

[math]a^x = T \left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -1}{\log_Ta -1}\right)}[/math]

 

is preferable to

[math]\frac{T}{T}a^x = T \left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -1}{\log_Ta -1}\right)}[/math]

 

. Whether you prefer

[math]\frac{a^x}{T} = \left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -1}{\log_Ta -1}\right)}[/math]

 

is largely aesthetic. I like it. I like even more this variation:

[math]Ta^x = (Ta)^{\left( \frac{x\log_Ta +1}{\log_Ta +1}\right)}[/math]

 

I don’t think it can be written much more tersely than this. :)

 

All of these are identities, with the restriction [math]T \not= a[/math], [math]T \not= 0[/math], [math]T \not= 1[/math], and [math]a \not= 0[/math].

[Don Blazys]

To:CraigD. If we begin with the term:

 

(T/T)a^x,

 

then "crossing out" the cancelled T's leaves us with:

 

a^x,

 

and if all we are left with is:

 

a^x,

 

then we can't possibly derive the more defined and powerfull term:

 

T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)).

 

That's one reason why I am against teaching students that they should automatically "cross out" cancelled common factors. I believe that students should be taught the truth, and the truth is that when confronted by a term such as:

 

(T/T)a^x,

 

they have a choice. They can either "lose" or "cross out" the cancelled T's and write:

 

a^x,

 

or, they can choose to retain the cancelled T's and write:

 

T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)),

 

where it is now algebraically impossible to "lose" or "cross out" the cancelled T's. Before I invented the above "cohesive term", we didn't have this choice!

 

Now, the question is, if a "common factor" is defined as a positive integer T>1, then which term should we use for problems that involve cancelled "common factors"...

 

a^x

 

or

 

T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) ?

 

I prefer the "cohesive term". How about you?

 

Don.

[Qfwfq]

Don, the little time I could afford when I read your post was wasted on making sure I had not mis-read, re-checking the parenthesis etc. Although I had not translated it wron onto paper, the effort in scarce time brought me to a hasty conclusion and only last evening (and because a sane person such as Craig confirmed your claim so I deemed the challenge worthwile) I scrounged the time at the risk of ruining my frugal supper and including the time to re-derive the logarithm rules after my initial senile muddle; at first I had written the ratio as the log of the difference...:hihi: ;)

 

So, yes it's an actual identity although not obvious and rather tricky to prove, but I don't see why it has profound implications about the basic methods of formal calculation.

 

Now I had at least granted you the benefit of doubt and, given my lack of time, invited you to support your own claim:

Don I see no way for the exponent to identically equal 1 independently of x and if there's any way of the logarithms in the exponent compensating the T's in the base, go ahead and show us.

but you did quite otherwise. This is not the way to increase your audience either; an effort to get along with people and to be more compliant would do better. :)

 

 

Egaaaaaaaaash!!!!! :) I saw it after!

That's one reason why I am against teaching students that they should automatically "cross out" cancelled common factors.

I agree with the word automatically and I now get what you're after but it's no new cool discovery! I don't however agree where you use the word wrong. I certainly wasn't taught to do these things automatically, indeed I was often taught that it's sometimes necessary to do the opposite of "crossing out". Formal computations can be like getting from A to B in the mountains, the way to go isn't always straight down the gradient... :hyper:

 

That's the difficult thing about learning to do exercizes and not many teachers are good at helping students to have the best approach to finding the way. My highschool math teacher was lousy in this and it caused me difficulty through the rest of my studies, but she definitely showed us how tricks are often necessary (add and subtract the same term, multiply and divide by the same term etc.). You seem to be banging on a wide open door instead of explaining your intent and helping to explore the complex inside of the palace.