I have a hard time with 'Natural Log's' too, Can someone please help explain it to me? I appreciate it :)
148470
To quote one of my old teachers, who made a lasting impression by screaming the following at our class:\n\nLOGRITHMS ARE EXPONENTS!!!!!!\n\nIf you have an exponent, then you are taking some number to a power.  That number is known as the base.\n\nYou can use any positive number for the base but usually log refers to log base 10. ln is always base e, where e = 2.7...\n\nThat is, saying ln(x) = y is the same thing as saying x = e^y. (y is a logrithm so y is an exponent!)\n\nHistory:\nNatural logs originate in the search for an integral for 1/x. \nIf you define a function, F(x), such that F'(x) = 1/x and start looking at its properties, you find that F(ab) = F(a) + F(b) and F(a^n) = nF(a). The behavior you would expect from a logrithm function. When you calculate the base for that function, you get e.
729730
u cant split a log like that! ie ln(ab) = lna +lnb\n\nu should use the f'(x) f(x)^n = f(x)^(n+1)/ (n+1)
499879
I'm trying to guess what the problems you were asked to do were, because the information you gave in posing the question was incomplete.\n\nIf f(x) = log(x)  (for log of any base)\n\nthen\n\nf(1/b) = log(1/b) = log(b^(-1))=-log(b)=-f(b)\n\n\nlog(MN^2) = log(M) + 2 log(N)\n\nor\n\nlog((MN)^2) = log(M^2) + log (N^2) = 2log(M) + 2log(N)\n\nThese are applications of the laws of logarithms.
100388
This is due to a property of log which is that :\nlog(ab)=log(a)+log(b)\nRemark that this is true when well defined (for eg you can't write log(-1*-3)=log(-1)+log(-3) )\n\nSo here, you have log(x^2)=log(x)+log(x)\n\nTherefore, there are no diff. except that log(x^2) exists for all x in R&#92;{0} and 2log(x) is defined for all x strictly positive.
205936
1) The natural logarithm started life as the integral of 1/t from 1 to x. This integral was defined to be the natural logarithm because it satisfied the properties of a logarithm in x. The base of this logarithm was called e, Euler's number. Describing the usefulness of this number, and thus why everyone accepted the term natural logarithm, fills several good books, but you can find a condensed version in the source link. \n2) A distribution which is log-normal may be written as a large sum over the logarithm of some random variable, and thus the logarithm of a product over that random variable. Consider the central limit theorem applied to the logarithm of a random variable. More information can be found in the source link below.
1035877
The natural logarithm (over the real numbers) is defined only for positive numbers, so \n\nx-4>0\n\nin other words\n\nx>4.
783096
yes, it really is easier than you think:\na^(log_a(2/x) +log_a(x))=a^(log_a(2) -log_a(x) +log_a(x))\n  =a^(log_a(2))\n  =2\nnote that i used the property log(a/b)= log(a)- log(b) and a^log_a(x)=x
704178
I'm sure you must have been given these few principles of log operations:\n1. log a + log b = log ab\n2. log a  - log b = log (a/b)\n3. log a^p          = p*log a\n4. log 1 = 0\n5. log (subscript b) b = 1\n\nIn the first problem, if you apply rule 3 it becomes\n1/5(log(x+1)^3 + log(x-1)^2 - log 7), then apply rules 1 and 2 and you get\n1/5log(((x+1^3)*(x-1)^2)/7), a horrible looking expression and I don't know why anyone would ask you to touch it!\n\nIn the second one, log (subscript b) x*b\n= log x + log b [by rule 1]\n= log x + 1     [rule 5]\nTo deal with log sqrt 10, remember \nsqrt 10 = 10^(1/2) and use rule 3;\nalso 10 = 2*5 and use rule 1 in reverse, i.e. split log 10 into log 2 and log 5.\n\nI can't do much more without actually doing them all for you, and I don't think I should do that.\n\nIn  problem 3, take the log of each side and use rule 3, i.e.\nto solve  a^(x + b) = c, do\nlog a^(x + b) = log c\nand then (x + b)*log a = log c\n\nYou can';t do q.4 without knowing what the base of logs is.  I'd assume it's 10 and use rule 1 and solve\nlog x(x+3) = log 10\n\nIn problem 7, I think those 4's are all subscripts, in that case you are right and have used the rule I omitted: \n\nlog (subscript b) x  = (log x)/(log b)\n\nHowever, if you apply rule 3 to log 64, using 64 = 4^3, you'll simplify your answer a great deal.  Is it 3?
437664
The secret is to understand some of the rules of logarithms and adjust translate the numbers to make your lob easier.\n\nThe rules:\nlog(a*b) = log(a) + log(b)\nlog(a/b) = log(a)- log(b)\nlog(a^n)=nlog(a)\n\nIf I understand your problem:\n\nx= [(86400)*sqrt(365.25)]/[ (.079)^11]\n\nso log(x) = log [.....]\n\n     = log(86400)+log(356.25^.5) - log(0.79^11)\n     = log(86400)+.5*log(356.25) - 11*log(0.79)\n\nuse log tables to get the logs and add them up to get the log of x.  Use the tables again to find x.\n\nYou can make other approximations, i.e., 356.25 is close to 19^2 (=361)\nand 86400 is close to 300^2 (=90000) and continue from there.  \n\nHope that helps.
1156569
The problem is not that you can't take the log of a complex number. The problem is that log is multiple valued in the complex plane. In other words, the exponential function is not one-to-one. You can take the 'principle branch' of the logarithm, but then you can lose certain of the rules of logarithms.\n\nThe upshot is that when you take the log of -1, you either get an infinite set of values (.., -pi*i,  pi*i, ...) and another infinite set of possibilities for the log of 1 OR you lose the property that log(a^b)=b*log(a) that you used in your first paradox. This is just a fact of life when working with complex numbers. In the second, BOTH -pi*i and pi*i are logarithms of -1, so the argument fails. It's somewhat like saying that the squares of -1 and 1 are the same, so -1=1.
965840
Can you write out in words what the above equation is supposed to be?  It looks like either a base of 2 or a base of 1/32 and a log of the other of those two.  Remember you can write log base a of b as log(a)/log(b) and that log(a^n) = n*log(a).  By those two properties, if the above is log base 2 of 1/32, then it is the same as log(1/32)/log(2) = log(1/(2^5))/log(2) = -5*log(2)/log(2) = -5.  Hope this helps, but clarify if this isn't what you were asking.
750081
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !
899135
1. Move the log part to the other side to make it positive, and move the 5 to the other side to get -2 = log(x+2)\nThe base of log is 10, so this means basically 10^-2 = x+2\n\n10^-2 is 1/100 or 0.01 = x + 2\n\n-1.99 = x\n\n2. 2e^(2x) + 5e^x - 3 = 0\n\nFactor by FOIL:  (2e^x - 1)(e^x + 3) = 0\n\nSo either e^x = 1/2 or e^x = -3 (which is impossible)\n\nYou can do ln (natural log) of both sides to "cancel out" the e:\n\nln(e^x) = ln (1/2) becomes x = ln(1/2)\n\n3. logx - 1/2log(x+4) = 1\n\nMinus becomes divide and the 1/2 becomes an exponent when combining logs.  So the left side simplifies to be:\n\nlog(x/(x+4)^(1/2))   Remember, 1/2 as an exponent means square root so this is\n\nlog(x/ sqrt(x+4)) = 1\n\nAgain, log has base 10 so this is really 10^1 = x/sqrt(x+4)\n\nSquare both sides: 100 = x^2/(x+4)\n\nCross multiply:  100(x+4) = x^2\n\n100x + 400 = x^2\n\nx^2 - 100x - 400 = 0  \n\nUse quadratic formula (be sure to eliminate negative answer(s) because you can't find the log of a negative number)
541208
it is a proving isn't it?\nright handside: [logbase10 1000]/logbase10 2\n=[logbase10 10^3]/[(logbase2 2 )/( logbase2 10)]\n=3[logbase10 10]x[logbase2 10]\n=3logbase2 10\n=logbase2 1000 , proved\n\n\n\nif its right do mark it best i'll come to know i have done correct or not.\nsorry here is the right answer\nanswer is 4\ni got it by\nlogbase2 1000=1ogbase10 1000 / logbase100 "x"\n3logbase2 10=31ogbase10 10 / logbase100 "x"\nlogbase2 10=1/logbase100 "x"\nlogbase100 10 / logbase100 2 =1/logbase100 "x"\n.5logbase100 100 / logbase100 2 =1/logbase100 "x"\n2logbase100 2=logbase100 "x"\nlogbase100 2^2=logbase100 "x"\n2^2="x"\n"x"=4
780572
First let's remember some rules of logs.\nThe log of product is the sum of the logs.  That is,\nlob(a*b) = log(a) + log(b)\nThe log of a quotitent is the difference of them.  That is, \nlog(a/b) = log(a) - log(b)\nAnd finally, if you have an exponent, you can bring it in front of the log.\nlog(a^n) = n*log(a)\nRemember that you can only take the log of positive numbers.\nln[ x^2 / [sqrt(x) (2+x)^2] ]\n= ln(x^2) - ln[ sqrt(x) (2+x)^2]\n= 2ln(x) - [ln(sqrt(x)) + ln(2+x)^2]\n= 2ln(x) - (1/2)ln(x) - 2ln(2+x)\nYou can combine 2ln(x) - (1/2)ln(x)\n2ln(x) - (1/2)ln(x) = (2 - 1/2)ln(x) = (3/2)ln(x)\nAnswer: (3/2)ln(x) - 2ln(2+x)\nNote: sqrt(x) = x^(1/2)
47515
There are lots of ways to find logarithms without a calculator. After all, people had to do this when there were no calculators. \n\n1. Logarithm tables are in books and all over the place. Here is an example of a log table on the internet:\n\nhttp://www.sosmath.com/tables/logtable/logtable.html\n\n2. My advisor in graduate school actually memorized a lot of values from the log tables so he could do very fast mental calculations. \n\n3. Slide rules are easy ways to get approximate logarithm values. \n\n4. There are numerical methods for approximating logarithms that you can track down, for various values of the argument. For example, look at \n\nhttp://www.maths.abdn.ac.uk/~igc/tch/ma1002/appl/node59.html\n\nhttp://mathworld.wolfram.com/NaturalLogarithm.html\n\nhttp://www.informatik.tu-darmstadt.de/ftp/pub/TI/reports/MarkusMaurer.diss.pdf\n\nAfter all, the calculator has to use some method to produce or approximate the logarithm, and all one has to do is to use similar methods to find the logarithm by hand, as long as they do not require an excessive amount of computing.
754734
1) The definition of a logarithm says : if a^X = N, then X = log[a]N.\nTurn this around to say : if X = log[a]N, then a^X = N.\nNow you can see what to do if we let X = log[2]64.\nWe get : 2^X = 64 = 2^6.\nTherefore, X = 6, that is, log[2]64 = 6.\n\n2) Same with this one. Let X = log[2]sqrt(2).\nThus, 2^X = sqrt(2) = 2^(1 / 2).\nTherefore, X = 1 / 2.\n\n3) 5^(-3) = 1 / 125\nUse the definition again. If a^X = N, then X = log[a]N.\nHere we have : a = 5, X = -3 and N = 1 / 125.\nThus, -3 = log[5](1 / 125)\n\n4) 10^(0.3010) = 2\nIn the same way, let a = 10, X = 0.3010 and N = 2.\nThus, 0.3010 = log[10]2.\n\n5) t = log[4]7\nDefinition again : 4^t = 7.\n\n6) log[10]7 = 0.845\nDefinition again : 10^(0.845) = 7.\n\n7) Let X = log[64]4\nBy the definiton : 64^X = 4.\nOr, (2^6)^X = 4.\nOr, 2^(6X) = 2^2.\nTherefore, 6X = 2, or, X = 1 / 3.