how the heck do you condense and expand logarithmic forms?
114069
log(a*b) = log(a)+log(b)\n\nlog(a/b) = log(a)-log(b)\n\nSo these work when expanding (left to right) and condensing (right to left).\n\n\nThanks for the update: So using your example:\n\n2ln3 - ln5 = ln(3^2) - ln5 = ln9 - ln 5 = ln(9/5)\n\nThe reason i converted 2ln3 to ln9 is because the condensing form only works when the coefficients of the logs are identical (ie 1 in this case). The formula for doing this is N*ln(x) = ln(x^N)\n\nNext, \n\nln(7X/3) = ln7 + lnX - ln3 (or ln(7/3) + lnX)\n\nAs you can see, the above form at 3 factors inside it - the expansion form will work for however many factors you want to use.\n\nIf you plug in the numbers into a calculator then you will see that this works.
943827
He should have put parenthesis around the 3X5^Y and then he was expanding the log(3X5^Y) to the log3 + log 5^Y because the sum of the logs is equal to the log of the products.  The log 5^Y can be written as Y(log 5).  You end up with\nY(log2) + log2     = log3 + Y(log5)\nY(log2) - Y(log5) = log3 - log2\nY(log2 - log5)      = log3 - log2\nY= (log3 - log2) / (log2 -log5)\nY= (log (3/2) / log(2/5) = log(1.5) / log(.4)\nY approximately = -.4425\nyou can check by substituting back in the original equation\n2^(-.4425 + 1) = 3 * 5^(-.4425) (use the reciprocal of 2nd factor\n2^.5575           = 3 / 5^(.4425)    because of the negative exponent)\n1.4717...         = 1.4717...
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
827167
log 2 = 0.3010(base10)
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.
734891
You can also use Google as a quick calculator.\nJust type in, say, log(6.54321)\nand the answer pops up as 0.815790859.
742533
Do you know your exponent rules? The same apply for logs. Multiplication means addition (this is why x^5 times x^2 is x^7 and NOT x^10) and division means subtraction (this is why x^16 divided by x^4 is x^12 NOT x^4).\n\nKeeping these rules in mind, I'll show you some examples of how these rules can be applied.\nlog x + log y = logxy\nlog 2x + log 4x = log8x^2\nlog12 = log6 + log2\nlog 4x - log 9x = log4x/9x\nlog2 - log4 = log2/4\nlog10x/3x = log10x - log 3x\n\n\nYou can also pull out numbers to the front of the logs like so:\nlog10x = 2log5x\nRemember with this rule: you can only pull NUMBERs out, NOT variables!\nSome more examples of this:\n4log4x = log16x\nlog45x = 9log5x\n2log6x = log12x\n\n\nNow I'll give you an example of all of them together:\n2log5x + 2log6x\nlog10x + log12x\nlog120x^2\n\nAnother example:\n4log4x - 2log2x\nlog16x - log4x\nlog16x/4x
658763
Take the logarithm of both sides:\n\nlog a^a = log 2\n\nThen use the rule log (x^y) = y log x\n\nso a log a = log 2\n\nUnfortunately from here we can only solve numerically. Look at the graph of a log a and make approximations.\n\nThough not expressible using elementary functions, the actual answer is defined by the Lambert's W function and reads: log 2 / (LambertW (log 2)), which is roughly 1.5596. More about the W function and what it's used for below.
575720
take logbase6 of 135 = x\n\nlog 135/log 6 = x\n2.737 = x\n\nThen substitute this into original to get:\n2^(log135/log6) + 1.3^(log135/log6)-2 \n\nIt should prove that it equals 30, which proves that both statements are ture.  However, I am getting that it is not equal to 30...so I think the proof is impossible,  unless your formula is typed incorrectly.
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.
290613
The sum of the logs is the log of the products, ie.\nlog(a)+log(b)=log(ab)\n\nSo combine the sum of logs here into a single log of the product of (x-3) and (x+3).\nlog_4(x-3)+log_4(x+3)\n  =log_4((x-3)(x+3))=2\n\nOnce you have a single log isolated on one side of the euqation you should remember that the log equation log_a(b)=x is equivalent to the exponential equation a^x=b, so in this case\n(x-3)(x+3)=4^2=16\n\nNow you've got a quadratic equation that you can solve for x by expanding everything, moving everything to one side of the equation, and factoring.\nx^2-6x+9=16\nx^2-6x-5=0\n(x-5)(x-1)=0\nx=1 or 5\n\nYou should then check to see if both of these solutions are valid in the original equation. Since letting x=1 would involve taking the log of a negative number in the original equation (which you cannot do if you are dealing with real numbers) you should throw that solution out.\n\nx=5
316037
METHOD 1:\n\nlog2[(6x-4)/(x+1)] = 4\n\nConvert to Exponential form:\n\n2^4 = (6x-4)/(x+1)\n\n16 = (6x-4)/(x+1)\n\nCross Multiply:\n\n16(x + 1) = 6x - 4\n16x + 16 = 6x - 4\n16x - 6x = -4 - 16\n10x = -20\n\nx = -2 ANSWER\n\nCheck:\n\nlog2[(6x-4)/(x+1)] = 4\nlog2[((6)(-2)-4)/(-2+1) ?=? 4\nlog2[(-12-4)/(-1) ?=? 4\nlog2[(-16)/(-1) ?=? 4\nlog2(16) ?=? 4\n4 = 4\n\nMETHOD 2:\n\nLog2[(6x-4)/(x+1)] = 4\n\nConvert to Base 10 Log. (Common Log.):\n\n(log[(6x-4)/(x+1)])/(log2) = 4\n\nCross Multiply:\n\nlog[(6x-4)/(x+1)] = 4(log2)\nlog[(6x-4)/(x+1)] = log(2^4)\nlog[(6x-4)/(x+1)] = log 16\n\nRemove the Logarithms:\n\n(6x-4)/(x+1) = 16\n\n6x - 4 = 16(x + 1)\n6x - 4 = 16x + 16\n6x - 16x = 16 + 4\n\n-10x = 20\n\nx = -2 ANSWER
885310
rt2logx+C\n\n2.sectheta+C\n\n3.is it [3^(2x)+1/x^(1/3)]dx?
41347
The trick to remembering the log rules is that they are the same as the exponent rules since logs are just rewritten exponents.\n\n(x^2)(x^3) = x^(2+3) = x^5\nlog [(a)(b)] = log a + log b\n\n(x^4)/(x^2) = x^(4-2) = x^2\nlog [a/b] = log a - log b\n\n(x^3)^2 = x^[(3)(2)] = x^6\nlog[a^b] = b[log a]\n\nThe rules for log and ln are the same since the only difference is log has base 10 (unless otherwise noted) and ln has a base of e.\nThere are also the two inverse functions with log and ln
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)
118410
...=(logA+logA)-(logB-logB-logB)+1/5logC=(logA^2-logB^3)+1/5logC\n= log(A^3/B^3)+1/5logC =log(A^3/B^3)+logC^1/5=log[(A^3*C^1/5)/B^3]\nNow if you know to transform C^1/5 (i don't really remember now)it becomes even simpler.
659018
Yes, there is a formula to change the base of a logarithm, and here it is...\n\nIf you have an expression log(x) with base 3, the change of base formula allows you to re-express this in the form (lg x)/(lg 3), where lg refers to a logarithm with a base of 10.\n\nFor information on the change of base formula and other logarithmic rules, you may find this page useful:--\nhttp://en.wikipedia.org/wiki/Logarithmic_identities