What is the best way to learn logarithms? Because I don't understand this in my Pre-Calculus class.?
106841
Try the following resources:\n\n1. http://www.krysstal.com/logarithms.html\n2. http://www.physics.uoguelph.ca/tutorials/LOG/\n1. http://en.wikipedia.org/wiki/Logarithms
916994
(log x)^3 = log (x^9)\n(log x)^3 = 9 log x\n(log x)^2 = 9\nlog x = +/- 3\nx = 10^3 = 1000  or x = 10^-3 = .001
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.
1187309
I think I did it.  If y=1/log(x), then a few manipulations gave me\n\nx=10^(1/y).\n\nI'm assuming by log you mean log to the base 10.  I suppose mathematicians will start fussing about what happens at y=0, but a physicist like me doesn't worry that much about infinities ...
132481
My old Calculus teacher had this pneumonic that was a good guide:\nL.I.P.E.T.\n\nFollow in the order of the letters to assign to u.  Look for first one in this list that your integral contains. Then assign that one to u:\nL-logarithmic functions\nI-inverse trigonometric funtions\nP-polynomial functions\nE-exponential functions\nT-trigonometric functions\n\nThen, assign everything that's left to dv.\nNext, take the derivative of u in respect to x, but move the dx over to the other side with x.\nThen, find v by integrating dv. For our purposes in finding v, leave off the constant C that you'd usually get for an indefinite integral.\nNext, use the following formula:\nIntegral of:(u dv) = uv - Integral of:(v du)\n-Where the Integral of:(u dv) is just your original integral.\n\nRepeat this process for the next integral if necessary.\n\nExample: \n("^" means raised to the power, e is the exponential constant)\n\nIntegral of:(xe^x dx)\nLet: u=x (x is a polynomial, so assign this to u; e^x is lower down on the list, so we don't use that for u)\ndu=dx\nLet: dv=e^x dx\nv=e^x\nIntegral of:(xe^x dx) = xe^x - Integral of:(e^x dx)\nIntegral of:(xe^x dx) = xe^x - e^x + C\n(Don't forget your constant C at the end)
871412
Because 10^1 = 10, you know that log(10) = 1.\n\nYou can simplify log(5) / log(10) to:\n= log(5) / 1\n= log(5).\n\nThe final part I can't do without a calculator... it's 0.69897.
914213
OK. It looks like you took the log base 10 of 9 (which is .9542) and multiplied it by 1/81 (which is .01234) to get your .01178. Thus, you found\n(1/81)*Log(9).\n\nTo get log base 9 of 1/81, i.e.\nlog_9 (1/81)\nyou have to find the exponent x where\n9^x=1/81.\nThis exponent will be -2 since\n9^(-2)=1/9^2=1/81.\nNo calculator is needed here!\n\nTo find the result on a calculator, you have to find\nLog(1/81) (which is -1.908485)\nand divide it by\nLog(9) (which is .9542425)\nto get -2.
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)
368201
(1/2)logx = log(x^(1/2)) = log(sqrt(x))\n\nso for exemple 1/2log(9)=log(sqrt(9)) = log(3).\n\nsometimes it's useful, sometimes it's more useful to write the log of a sqrt as 1/2 the log of the argument.
329438
log_5(X) + log_5(x-1)=log_5(6)\nlog_5(x(x-1)) = log_5(6)\nx(x-1) = 6\nx^2 - x - 6 = 0\n(x-3)(x+2) = 0\nx = 3 or x = -2\nYou cannot take the log of a negative number, so -2 cannot be a solution.\nAnswer: x=3\nFor the second part, find T(30) and T(75)\nT(30) = 70 + 100e^(-0.04(30))\n= 70 + 100(.3012)\n= 70 + 30.12\n= 100.12\n\nT(75) = 70 + 100e^(-.04(75))\n= 70 + 100(.0498)\n= 70 + 4.98\n= 74.98
851770
take the log\nso log b= log (1+(r/n))^n*t\nlog b = nt log (1+(r/n))\n{(log b)} / {log (1+(r/n))} = nt\n[{(log b)} / {log (1+(r/n))}] / n = t\ni hope those parenthesis dont mess u up lol\ni just learned this in school
362293
The logarithm (see attached article 1) of a number x in base b is the number n such that b^n = x. It is usually written as log_b(x) = n (b in a subscript).\n\nFor example, log (100) = log_10 (100) = 2, because 10^2 = 100.\n\nThere is one special case - natural logarithm (in base e =~ 2.7182818). It has some properties, which are very useful in calculus. They are written as ln (x).\n\n\nAntilogarithm is a synonym of exponentiation or power.\nThis is an inverse function to logarithm.\nIf log_b(x) = y, then b^y=x.\nFor integer x, b^x mean multiplication b*b*b*...*b (x times).\nFor all real x, b^x are defined in the second attached article, part "Exponentiation with real exponents".\n\nHope it helps.
694278
Log(x + 2) + Log(x - 1) = Log(70)\nAs I explained before,\nLog (a0 + Log (b) = Log(ab)\nSo,\nLog(x + 2) + Log(x - 1) = Log [(x + 2)(x - 1)]\nLog [(x + 2)(x - 1)] = Log(70)\n(x+2)(x-1) = 70\nx^2 - x + 2x - 2 = 70\nx^2 + x = 72 \nx^2 + x = 72\nx^2 + x - 72 = 0\nx^2 - 8x + 9x - 72 = 0\nx(x - 8) + 9(x - 8) = 0\n(x + 9)(x - 8) = 0\nNow, either x + 9 = 0 or x - 8 = 0\nIf we put x + 9 = 0, we get x = -9\nIf we put x - 8 = 0, we get x = 8\nBut when we substitute x = -9 in the original problem we get Log (-7) + Log (-10) = Log(70)\nBut Log of negative numbers if undefined.\nSo, x = 8 is the only solution.
862518
When we need to ind the logarithms of number not of base ten, we can use log*6*x=logx/log6 so the answer is\n(log60)/(log6)-(log30)/(log6)
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.
999317
(1) Simple inverse is log X as long as log is defined for the specific value of X, except that it's also not defined at X=1 since log(1) = 0.\n(2) Inverse function: If Y = 1/logX, then X = exp(1/Y) assuming natural logarithms. This is defined except at Y=0, which can't hold for finite X.\n\nThe proofs of both these are trivial. They are obvious.
643371
First you need to know what logarithms are. (logs)\nAll numbers can be written as a power of ten.  These are logs.  \nUsing a base of ten, \nlog 100 = 2 (because 100= 10 x 10 = 10 to the power of 2)and \nlog 1000 = 3 (because 1000= 10 x 10 x 10 = 10 to the power of 3).  \nNumbers between 100 and 100 will have a log between 2 and 3.  You can find the power by pressing the log button on the calculator.  Often people use natural logs which are powers of e (2.71...). For these press ln\nAnti-logs are the original numbers that the logs represent. Anti-logging is the process of finding the number represented by the log.  These days you would use a calculator and press inv log.  \nFor example log 50 = 1.70\nantilog 1.70 = 50\nwhich is the same as 10 to the power 1.70 = 50.\nIn the olden days, 25 years ago, before calculators Logs were useful when multiplying large numbers. You would find the logs of the numbers and then add the logs, before antilogging the answer to find the result of the multiplication. By subtracting logs you can do division. \n\nNow, just get out the calculator. I haven't heard the term antilog for many years as you would better represent your working using log and then using powers.