MWresearch Posted September 16, 2015 Share Posted September 16, 2015 (edited) Just what the title says. Say for instance I have [math]ln(x^2/3)[/math]. I want to be able to turn that into [math]2ln(x...[/math]But, I'm not sure exactly what's happening to a number or what the exact operation is when you take an exponent out of the argument of [math]log(x)[/math]. So, I'm not sure exactly what the result is if I assume [math]ln(x^2/3)=2ln(x/sqrt(3))[/math] because every time I try to do something like that, I get some complex constant left over. It "seems" like when you move exponents out of the argument of a logarithm, you're exponentiation everything in the argument by 1/exponent-you-want-to-get-rid-of and for some random reason that exponent then gets moved to the outside of the log operator, but then again, there's always a constant left over when I do that. I guess I could limit it to [math]x*a>0[/math] for [math]log(x*a)[/math] but I'd prefer to have the whole picture of whatever that constant comes from like [math]log(e^(ip*a)))[/math] or whatever. Btw the exponent command for the [math tags are completely broken and useless. Edited September 16, 2015 by MWresearch Link to comment Share on other sites More sharing options...

Strange Posted September 17, 2015 Share Posted September 17, 2015 You need to think about how logs work. Addition in "log space" becomes multiplication of the numbers; i.e. [math]\log(a) + \log(b) = \log(a * b)[/math]. Multiplication of logs becomes powers of the numbers: [math]n\log(a) = \log(a^n)[/math]. Reversing that last one seems to describe what you are trying to do. Combining these we get [math]\log(a^n * b^n)= \log((a * b)^n) = n \log(a * b) = n (\log(a) + \log(b))[/math] Btw the exponent command for the [math tags are completely broken and useless. They work fine. [ math]\log(e^(ip*a))[/math] => [math]\log(e^(ip*a))[/math] [ math]\log(e^{ip*a})[/math] => [math]\log(e^{ip*a})[/math] Note the use of {} to group things in Latex. 1 Link to comment Share on other sites More sharing options...

MWresearch Posted September 17, 2015 Author Share Posted September 17, 2015 That doesn't really explain anything, that's just a reiteration of basic basic basic logarithm rules that most people here learned years and years and years ago. What I'm talking about is, what the hell is going on when that happens? And that has to at least be related to complex analysis and Euler's Identity because of the constant left over in the negative portion of the domain of the argument of log(a*x). I'm looking for a formula that works for all arguments of log(x). -2 Link to comment Share on other sites More sharing options...

mathematic Posted September 18, 2015 Share Posted September 18, 2015 [latex]ln(\frac{x^2}{3})=2ln(x)-ln(3)[/latex] Link to comment Share on other sites More sharing options...

imatfaal Posted September 18, 2015 Share Posted September 18, 2015 [latex]ln \left( \frac{a \cdot x^j}{b \cdot y^k} \right) = ln(a) - ln(b) + j \cdot ln(x)- k \cdot ln(y)[/latex] The above variables are positive. Link to comment Share on other sites More sharing options...

MWresearch Posted September 18, 2015 Author Share Posted September 18, 2015 [latex]ln \left( \frac{a \cdot x^j}{b \cdot y^k} \right) = ln(a) - ln(b) + j \cdot ln(x)- k \cdot ln(y)[/latex] The above variables are positive. Ok so you keep saying x>0, but that's exactly what my issue is. I don't care about those basic rules for positive real numbers, the problem is what's going on that makes the result such that the above statements aren't true for negative numbers. Clearly, factoring out exponents for positive numbers is just "pattern recognition," not the whole story. Link to comment Share on other sites More sharing options...

imatfaal Posted September 18, 2015 Share Posted September 18, 2015 Ok so you keep saying x>0, but that's exactly what my issue is. I don't care about those basic rules for positive real numbers, the problem is what's going on that makes the result such that the above statements aren't true for negative numbers. Clearly, factoring out exponents for positive numbers is just "pattern recognition," not the whole story. Isn't it amazing that every thread people just answer the wrong questions. Nothing to do with the fact that maybe you are rambling and imprecise in your OP and intolerant and waspish in replies. A negative value is easily dealt with as I think you might have been getting at in the OP [latex] ln (-x) = ln (-1 \cdot x) = ln (e^{\pi \cdot i} \cdot x) = ln(e^{\pi \cdot i}) + ln(x) = \pi \cdot i + ln(x) [/latex] Link to comment Share on other sites More sharing options...

uncool Posted September 18, 2015 Share Posted September 18, 2015 (edited) Isn't it amazing that every thread people just answer the wrong questions. Nothing to do with the fact that maybe you are rambling and imprecise in your OP and intolerant and waspish in replies. A negative value is easily dealt with as I think you might have been getting at in the OP [latex] ln (-x) = ln (-1 \cdot x) = ln (e^{\pi \cdot i} \cdot x) = ln(e^{\pi \cdot i}) + ln(x) = \pi \cdot i + ln(x) [/latex] Err, it's not as easy as that. Because that method either gives multiple outputs for a given input, or breaks some of the usual laws we like to keep. Is the above equation true for all x? If so, then [LATEX]0 = ln(1) = ln(-(-1)) = \pi \cdot i + ln(-1) = \pi \cdot i + \pi \cdot i + ln(1) = 2 \pi \cdot i + 0 = 2 \pi i[/lAtEx] Is it only true when x is positive? If so, then it breaks logs of products, as we should have that [LATEX]0 = ln(1) = ln((-1)*(-1)) = ln(-1) + ln(-1) = 2 \pi \cdot i[/LATEX]. So it's not as easy as you're saying. The best answer is probably to allow a multivalued output; next best is probably to deal with a quotient of the complex numbers. Edited September 18, 2015 by uncool Link to comment Share on other sites More sharing options...

imatfaal Posted September 18, 2015 Share Posted September 18, 2015 Err, it's not as easy as that. Because that method either gives multiple outputs for a given input, or breaks some of the usual laws we like to keep. Is the above equation true for all x? If so, then [LATEX]0 = ln(1) = ln(-(-1)) = \pi \cdot i + ln(-1) = \pi \cdot i + \pi \cdot i + ln(1) = 2 \pi \cdot i + 0 = 2 \pi i[/lAtEx] Is it only true when x is positive? If so, then it breaks logs of products, as we should have that [LATEX]0 = ln(1) = ln((-1)*(-1)) = ln(-1) + ln(-1) = 2 \pi \cdot i[/LATEX]. So it's not as easy as you're saying. The best answer is probably to allow a multivalued output; next best is probably to deal with a quotient of the complex numbers. In rough and ready practice you can treat the ln(-x) as equal to the ln(x) +i.pi - It is the output of the complex logarithm function when the imaginary component is zero. And yes the complex logarithm function can have many answers with any multiple of 2.i.pi being added in turn . The answer to your simple question is ln(1) is zero - when you ask about -1 you are using complex logarithms and the principle value is 0 but any integer multiple of 2.i.pi + 0 will also be a logarithm Link to comment Share on other sites More sharing options...

MWresearch Posted September 18, 2015 Author Share Posted September 18, 2015 (edited) A negative value is easily dealt with as I think you might have been getting at in the OP [latex] ln (-x) = ln (-1 \cdot x) = ln (e^{\pi \cdot i} \cdot x) = ln(e^{\pi \cdot i}) + ln(x) = \pi \cdot i + ln(x) [/latex] I tried explaining it and instead of taking my advice on what I'm asking about you decided to act inappropriate and then on top of it you still create the exact same issue but for negative numbers. No separation, there should be one formula for a set of all real numbers if not all complex numbers z, a formula where not only can both positive and negative numbers correspond to ln(x) and ln(-x) but also give the +2pi*i*n for all other possible solutions that result from log(x)'s trigonometric relation. Edited September 18, 2015 by MWresearch -3 Link to comment Share on other sites More sharing options...

mathematic Posted September 18, 2015 Share Posted September 18, 2015 Ok so you keep saying x>0, but that's exactly what my issue is. I don't care about those basic rules for positive real numbers, the problem is what's going on that makes the result such that the above statements aren't true for negative numbers. Clearly, factoring out exponents for positive numbers is just "pattern recognition," not the whole story. [latex]ln(x^2)=2ln(|x|) [/latex] for real x. Link to comment Share on other sites More sharing options...

MWresearch Posted September 19, 2015 Author Share Posted September 19, 2015 (edited) [latex]ln(x^2)=2ln(|x|) [/latex] for real x. And that explains the complex result of a negative argument in addition to the real values for all positive and negative values of the argument in one formula how????? I don't know how many times I have to explain it but I'll do it until you get it. The formula shouldn't restrict anything to only positive values which abs does. Somewhere out there, there's a single extended form of log(x) that explains not only what's happening when you move the exponent but what the complex constant is added for negative values in the formula that simultaneously accounts for both the positive values of the argument yielding only real numbers while the negative values of the argument yield complex numbers and that's what this is all about. Every logarithmic rule that people have written so far only works for either only positive inputs or only negative inputs, but, it should be both for the continuum of results. Edited September 19, 2015 by MWresearch Link to comment Share on other sites More sharing options...

mathematic Posted September 19, 2015 Share Posted September 19, 2015 I don't understand your concern - there are problems for the general case. However for this case [latex]x^2\ge 0[/latex] for all real x, so the expression ([latex]ln(x^2)=2ln(|x|)[/latex]) holds, because if x is negative, that information is lost after squaring. 1 Link to comment Share on other sites More sharing options...

MWresearch Posted September 20, 2015 Author Share Posted September 20, 2015 (edited) I don't understand your concern - there are problems for the general case. However for this case [latex]x^2\ge 0[/latex] for all real x, so the expression ([latex]ln(x^2)=2ln(|x|)[/latex]) holds, because if x is negative, that information is lost after squaring. Except you're not considering the arbitrary coefficient "a" and still completely ignoring that I keep referring to possible positive and negative values in general, not solely x^2. What if I had [math]-x^{2}[/math]? Then all values would be negative. There should be one general equation that explains all outputs of log(x) that tells me what the complex numbers are for negative arguments along with their continuum of 2*pi*n solutions resulting from it's trigonometric relation through Euler's identity as well as simultaneously giving me only real output values when the argument is positive, no constraints like x>0 or x<0 for any argument. Basically, an extension of the log(x) function which by hand one could derive the correct output values all real values of x, not only positive values and not only negative values, all values. Edited September 20, 2015 by MWresearch Link to comment Share on other sites More sharing options...

swansont Posted September 20, 2015 Share Posted September 20, 2015 there should be one formula for a set of all real numbers ! Moderator Note You don't get to demand that. The logarithm of a negative number is undefined, so you're stuck with it. Getting pissy about it doesn't change anything. 2 Link to comment Share on other sites More sharing options...

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