Don Blazys 10 Posted September 11, 2008 Share Posted September 11, 2008 Is the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) true or false? If true, then can we "cross out" the T's and make them "disappear" the way we were taught in school? Don. Link to post Share on other sites

ydoaPs 1601 Posted September 11, 2008 Share Posted September 11, 2008 AFAICT, the answer for T would make the equation indeterminant. Link to post Share on other sites

Don Blazys 10 Posted September 11, 2008 Author Share Posted September 11, 2008 (edited) Since unity can be defined as 1=(T/T) where T>1, the answer for T is not indeterminate. This is a simple true/false, yes/no question. Don. Edited September 11, 2008 by Don Blazys Link to post Share on other sites

the tree 222 Posted September 11, 2008 Share Posted September 11, 2008 Trying to wade through that mess of parenthesis, I think the equation that I'm looking at is [math](\frac{T}{T})a^x=T(\frac{a}{T}) ^{ x \frac{\frac{\ln{a}}{\ln{T}} -1 }{ \frac{\ln{a}}{\ln{T}} -1 }}[/math] which is clearly true. Aside, what importance does this have? Or was that some sort of a joke? Link to post Share on other sites

Don Blazys 10 Posted September 12, 2008 Author Share Posted September 12, 2008 You are correct, (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is indeed a true equation (identity). Obviously, it would remain true even if we somehow transported the right hand side all the way to the other side of the universe. Now, can we "cross out" the T's and make them "disappear" the way kids are being taught to do in school? Don. Link to post Share on other sites

Bignose 946 Posted September 12, 2008 Share Posted September 12, 2008 (edited) Trying to wade through that mess of parenthesis, I think the equation that I'm looking at is [math](\frac{T}{T})a^x=T(\frac{a}{T}) ^{ x \frac{\frac{\ln{a}}{\ln{T}} -1 }{ \frac{\ln{a}}{\ln{T}} -1 }}[/math] which is clearly true. No, it isn't clearly true. Firstly, to Don, yes anything divided by itself is equal to one. "Making them disappear" is normally called "canceling" but yes, when the exact same factor appears in the numerator and denominator, they can just "disappear". Now, let's look at [math](\frac{T}{T})a^x=T(\frac{a}{T}) ^{ x \frac{\frac{\ln{a}}{\ln{T}} -1 }{ \frac{\ln{a}}{\ln{T}} -1 }}[/math] Taking that as correct (as verified by Don), the numerator and denominator in the exponent on the right hand side at the same. [math]\frac{(\frac{\ln{a}}{\ln{T}} -1) }{ (\frac{\ln{a}}{\ln{T}} -1) } = 1[/math] because both the numerator and denominator are the same. So the big equation becomes: [math](\frac{T}{T})a^x=T(\frac{a}{T}) ^{ x }[/math] The above equation simplifies to [math]a^x = T^{1-x}a^x[/math] Which clearly isn't "true", at least not always. It it true only for three special cases: 1) T=1 2) x=1 or 3) a=0 and x!=0. Edited September 12, 2008 by Bignose Link to post Share on other sites

Don Blazys 10 Posted September 12, 2008 Author Share Posted September 12, 2008 To: Bignose. The equation is true. The expressions involving logarithms "cancel out" if and only if x=1. Otherwise, they don't. Don. Link to post Share on other sites

DrP 591 Posted September 12, 2008 Share Posted September 12, 2008 To: Bignose. The equation is true. The expressions involving logarithms "cancel out" if and only if x=1. Otherwise, they don't. Don. Why should the value of x matter? [(lna/lnT) -1] / [(lna/lnT) -1] = 1 regardless of the value for x, surely? Link to post Share on other sites

Dave 251 Posted September 12, 2008 Share Posted September 12, 2008 What is the point of this thread? Link to post Share on other sites

the tree 222 Posted September 12, 2008 Share Posted September 12, 2008 No, it isn't clearly true. ... [math](\frac{T}{T})a^x=T(\frac{a}{T}) ^{ x }[/math] The above equation simplifies to [math]a^x = T^{1-x}a^x[/math] Obviously I was being careless, you are of course right. What is the point of this thread?lulz? Link to post Share on other sites

Dave 251 Posted September 12, 2008 Share Posted September 12, 2008 lulz? As much as I admire the lulz, this thread contains no lulz. Link to post Share on other sites

Dr. Jekyll 10 Posted September 12, 2008 Share Posted September 12, 2008 Trick question: What if [math]T=a[/math]? Quite surprised none even mentioned that (it was my first thought), maybe I overlooked some defs.. Link to post Share on other sites

Bignose 946 Posted September 12, 2008 Share Posted September 12, 2008 Trick question: What if [math]T=a[/math]? Quite surprised none even mentioned that (it was my first thought), maybe I overlooked some defs.. If T=a, then the equation is only true if T=a=1 or x=1. -------------------------- But, I want to second dave's question here: What is the point of the thread? It needs to be at least re-titled from "The Most Important Question Ever" to "please check my cancellation". Because, there is nothing important about this equation at all right now, and the OP hasn't attached any significance to it at all. To: Bignose. The equation is true. The expressions involving logarithms "cancel out" if and only if x=1. Otherwise, they don't. Don. If x = 1, then there is no point in even leaving x in the equation, and you end up with the very unprofound result a=a. I don't think that a=a is "The Most Important Equation Ever", no matter what a turns out to be. Link to post Share on other sites

Dr. Jekyll 10 Posted September 12, 2008 Share Posted September 12, 2008 If T=a, then the equation is only true if T=a=1 or x=1. Indeed! I don't think that a=a is "The Most Important Equation Ever", no matter what a turns out to be. Hahaha . However, everyone who has tried to prove something has at one time or another ended up with 0=0, 1=1 or such. To defend the poster, I think it is a very important equation since it tells us we have ****ed up! Link to post Share on other sites

Don Blazys 10 Posted September 13, 2008 Author Share Posted September 13, 2008 To: Dr.P. The exponent that I wrote is ((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)). Don't confuse it for x((ln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)). Don To: Dave, Bignose and Tree. The point of this thread is that if the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is true, then even though the T's cancel, they can not be "crossed out" and rendered "invisible" in the way that is taught in our schools. Moreover, letting T=1 demonstrates that multiplication and/or division by unity automatically results in division by zero! Don. To: Dave and the Tree. What does "lulz" mean? Link to post Share on other sites

the tree 222 Posted September 13, 2008 Share Posted September 13, 2008 To: Dr.P. The exponent that I wrote is ((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)). Don't confuse it for x((ln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)). DonI did say I think. The LaTeX feature is available to you to and it'd be nice if you would use it. To: Dave, Bignose and Tree. The point of this thread is that if the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is true,Well, it isn't. then even though the T's cancel, they can not be "crossed out" and rendered "invisible" in the way that is taught in our schools.Crossing out is a synonym of cancelling. Moreover, letting T=1 demonstrates that multiplication and/or division by unity automatically results in division by zero! Don.No it doesn't. Link to post Share on other sites

Don Blazys 10 Posted September 13, 2008 Author Share Posted September 13, 2008 To: Dr. Jekyll and Bignose. (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is a proven identity. Therefore, it must be true for all possible values of the variables. The special case T=a requires x=1 because if Ta^x is a composite number and T is the greatest possible factor, then (T/T)a^x must represent a prime, and primes can't be represented as having exponents. Don. To: Tree. First you said that the equation is true, and now you are saying that it's false. You were right the first time! Don. Link to post Share on other sites

the tree 222 Posted September 13, 2008 Share Posted September 13, 2008 To: Tree. First you said that the equation is true, and now you are saying that it's false. You were right the first time! Don.I was mistaken the first time and I have already conceded that BN corrected me. Besides, you'll note that I was not originally looking at the equation correctly. The equation does in fact hold except for certain cases which, an identity, it does not make. Link to post Share on other sites

Don Blazys 10 Posted September 13, 2008 Author Share Posted September 13, 2008 To: Everyone who reads this. It's election time, so let's have a vote. Is the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) true or false? No commentary please, just vote true or false over the weekend. I will be back on Monday to count the votes. Don. I will cast the first vote. I vote TRUE. Don. Link to post Share on other sites

Bignose 946 Posted September 13, 2008 Share Posted September 13, 2008 (edited) (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) Don, for this complicated of an equation, please, please, please use LaTeX -- it is much clearer. Or even just break it down. Write T(a/T)^r where r = s/t and s = .... and t = .... All those ( and )'s are really confusing. I think that this is closer to what Don intended. [math](\frac{T}{T})a^x=T(\frac{a}{T}) ^{ \frac{\frac{x\ln{a}}{\ln{T}} -1 }{ \frac{\ln{a}}{\ln{T}} -1 }}[/math] But I cannot be sure. I'm not going to be sure unless he posts something clearer. nevertheless the first term T/T can be cancelled out whether Don thinks so or not. So the LHS is just [math]a^x[/math]. We can do some simplifying to the RHS, too. First, I'd multiply the exponent in the RHS by ln(T)/ ln (T) in both the numerator and denominator to clean up some fractions. [math] \frac{\frac{x\ln{a}}{\ln{T}} -1 }{ \frac{\ln{a}}{\ln{T}} -1 } = \frac{ x\ln{a}-\ln{T} }{\ln{a}-\ln{T}} [/math] Next, I'd use some of the properties of logarithms to clean up the -'s and +'s. [math] \frac{ x\ln{a}-\ln{T} }{\ln{a}-\ln{T}} = \frac{ \ln{ \frac{a^x}{T} } }{\ln{\frac{a}{T}}} [/math] That's about the best I can see. Maybe there is more. So, finally, I end up with [math] a^x = T(\frac{a}{T}) ^ {\frac{ \ln{ \frac{a^x}{T} } }{\ln{\frac{a}{T}}}} [/math]. This equation is not going to be true in a general sense. I don't even think that it is true for any special cases -- at least none that I can spot quickly. The [math]\ln{a}[/math] term in the exponent in RHS really mucks things up. Oh, and science and math is NOT, NOT, NOT a democracy. So, I'm not going to "vote". This equation IS FALSE -- with the possible exception of a few special cases that I haven't taken the time to look for. In general, it is definitely FALSE. Any chance we're going to actually learn the motivation behind this thread any time soon? Edited September 13, 2008 by Bignose Link to post Share on other sites

the tree 222 Posted September 13, 2008 Share Posted September 13, 2008 This equation is not going to be true in a general sense. I don't even think that it is true for any special cases -- at least none that I can spot quickly.Have you tried throwing some random numbers in?Here's a sample of some Maple output ex := simplify(%); x ln(a) - ln(T) --------------- ln(a) - ln(T) T*(a/T)^ex; /x ln(a) - ln(T)\ |---------------| \ ln(a) - ln(T) / /a\ T |-| \T/ eq := simplify(%); /x ln(a) - ln(T)\ |---------------| \ ln(a) - ln(T) / /a\ T |-| \T/ a := 2; T := 3; x := 4; 2 3 4 a^x; evalf(eq); 39.0625 39.06249990 a := 7; T := 2; x := 3; 7 2 3 a^x; evalf(eq); 343 343.0000002 I can't decide whether those are small differences or rounding errors, and given the lack of motivation I don't care much either. But the results do seem to correlate. Obviously proof would take precedence over experiment. Link to post Share on other sites

Josy 10 Posted September 13, 2008 Share Posted September 13, 2008 Any chance we're going to actually learn the motivation behind this thread any time soon? AFAICT the point is that if the equation in question held in general, it would prove that cancellation of common factors doesn't work. Since it doesn't hold, it pretty clearly proves nothing of the sort. This is unsurprising, since cancellation arises very fundamentally from the way the real numbers work, and is pretty obviously correct. Link to post Share on other sites

Bignose 946 Posted September 13, 2008 Share Posted September 13, 2008 Well, I've changed my mind. I did some pencil & paper work here and found how they are indeed equal. I couldn't see it just from looking at it, I had to do a bunch of chicken-scratch work. The extra T in there throws off the simplify function in MathCAD and Maple, too. I don't think that it needs to be posted, if it does, the OP should probably do so. But, ultimately, I still don't see a point. Don, is there a point? Needlessly making something more complex isn't really that big of a deal. I mean, we could write [math]-e^{i\pi}[/math] every time we wanted to write 1, but what is point? I don't see why "cancellation of common factors doesn't work" just because the equation he presented is true. I actually used cancellation of common factors several times in my derivation of it. The equation can be shown to be true precisely because cancellation of common factors works. So, I really don't see the point. Sorry. I am still hoping Don will explain it, however. Link to post Share on other sites

Don Blazys 10 Posted September 16, 2008 Author Share Posted September 16, 2008 To: Bignose. Congratulations! You got it! Sorry about not using "LaTex" but I never even touched a personal computer until a few months ago and my "computer skills" are, at this time, limited to checking and sending e-mails, performing "Google searches" and posting in forums. I'm almost 60 years old and "early onset alzheimers" makes it difficult to learn and retain new skills. You know, when I was in high school back in the 60's, we didn't have personal computers, calculators, or any of that stuff. I still use a "slide rule", pencil and paper... and for most mathematicians my age, parenthesis make equations easier, rather than more difficult to understand. In other words, I wasn't trying to make things difficult for you. Perhaps you would be kind enough to post the formal derivation of my equation in "LaTex" in your next post on this topic. That way, this new generation of mathematicians that relies on the "high tech" that my generation invented will find my work easier to understand. You can find that derivation on my website (donblazys.com). There, you will also find the point of this topic. Don. To: Josy. (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is a proven identity, but it is very very "counterintuitive" because even though the T's cancel (are neutralized) on both sides, they can't be "crossed out" on either side because if we "cross out" the T's on the left, then we have no way to logically derive the term on the right, where the T's can"t be "crossed out". Don't give up! You were right when you said that it shows the present method of eliminating common factors to be inadequate. By the way, you can find more information about this equation on my website, and on the Marilyn Vos Savant forum. Thanks for your interest. Don. To: The Tree. I predict that you will change your mind yet again. Good luck! Don. Link to post Share on other sites

DrP 591 Posted September 16, 2008 Share Posted September 16, 2008 . (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is a proven identity, but it is very very "counterintuitive" because even though the T's cancel (are neutralized) on both sides, they can't be "crossed out" on either side because if we "cross out" the T's on the left, then we have no way to logically derive the term on the right, where the T's can"t be "crossed out". I still don't get it Don. If we cross out the T's on the LHS side we are left with: a^x = RHS - - so what? I don't see the problem - can you explain it to me please? Link to post Share on other sites

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