What's a good approximation to calculate log(n!)?
543595
Stirling's approximation:\n\nn! is approximately\n\nsqrt(2*pi*n)*(n/e)^n,\n\nso ln(n!) is about\n(n+1/2)*ln(n)-n +(1/2)ln(2pi)
790275
log(9)3\n=\nlog(9)9^1/2\n=\n1/2*log(9)9\n=\n1/2\n[as log(m)m=1]
819937
You haven't put brackets in, so the question looks ambiguous.\n\nIt could be [log(x)] ^ [log(x)], which is difficult,\nor it could be log[x ^ log(x)], which is easier.\n\nI'll take the easy way out.\n\nLet log[x ^ log(x)] = 25, where I'll assume log is to the base 10.\n\nBy one of the rules of logarithms, which states that\nlog(M ^ N) = N * log(M), we have :\n\nlog(x) * log(x) = 25\nor, [log(x)] ^ 2 = 25\n\nTaking the square root of both sides gives :\nlog(x) = ± 5\n\nAs I've assumed this is to base 10, this gives :\nx = 10 ^ (± 5)\nSo, x = 10 ^ 5 or 10 ^ (-5),\nwhich is 100,000 or 0.00001.
1140302
Here is an interactive page that discusses one way to do it:\n\nhttp://dilc.upd.edu.ph/loDeploy/math/math10a.html\n\nFor large numbers, bequalmings technique using logs is an easier way to go.
1093339
No one on here is going to actually DO your homework for you.. but here is a place for you the get help -\n\nhttp://www.themathpage.com/aPreCalc/logarithms.htm\n\nhttp://en.wikipedia.org/wiki/Logarithm\n\nhttp://physics.mtsu.edu/~wmr/log_2.htm
324025
log 10000 = x\n10^x = 10000\nx = 4\n\n4
789068
if this log base 6 of 60 and 30 or log base 10 of 660 and 630?\n\nif it's the first \n\nlog6 60 is log60/log6\nlog6 30 is log30/log6\n\nplug log60/log6 - log30/log6 into your calculator!\n\nif it's the later, plug log660 - log630 into your calculator. if you don't have one, try http://www.math.com/students/calculators/calculators.html
246299
h= (C*I/A+T/k)*z*log(.7*y)
881778
(1.0442) to the power n(number of years)=1.7, the ratio of the final to the original amounts. \nFrom logarithms, n log (1.0442) = log 1.7. considering the same base (let us say 10).\nlog 1.7= 0.2304489214\nlog 1.0442= 0.017867719\nhence, log1.7/log1.0442= 12.2685657176, say approximately in 12 years.
812579
Adding logs is the same as taking the log of the two numbers multiplied. \nSubtracting the logs is the same as taking the log of the ratio (left number divided by the right number).\nlog (a) 15 - the log (a) 3 = log (a) 15/3 = log (a) 5.
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?
689473
get the n's on one side\nso\n17959/21558=(1.02/1.03)^n\nor\n.833055014=.990291262^n\nusing logs with change of base\nn=log (.833055014)/log(.990291262)\nor n=18.72
958491
i think we can convert the log with subsrcipt 3 to natural log\nlogx (with subscript 3) = lnx/ln3\nso derivative is\n1/(xln3)\nhope this helps
867514
[ 1/4 LOG (16) ] - [ 2 LOG (6) ] + [ LOG (3) ]\n\n[ LOG ((16)^(1/4)) ] - [ LOG ((6)^2) ] + [ LOG (3) ]\n\n[ LOG (2) ] - [ LOG (36) ] + [ LOG (3) ]\n\n...........................the first two LOGS have the same base of 10, \n...........................so when subtracting LOGS we divide.\n\n[ LOG (2/36) ] + [ LOG (3) ]\n\n[ LOG ((2divided by 2) / (36 divided by 2)) ] + [ LOG (3) ]\n\n[ LOG (1/18) ] + [ LOG (3) ] \n\n...........................the above two LOGS have the same base of 10, \n...........................so when adding LOGS we multiply.\n\nLOG ((1/18) times (3))\n\nLOG ((1/18) times (3/1))\n\nLOG (3/18)\n\n..................................then reduce the fraction by dividing by 3/3.\n\nLOG ((3 divided by 3) / (18 divided by 3))\n\nLOG (1/6)   \n\nLOG ((6)^(-1))\n\n..........an exponent can be moved and multiplied to the front of the LOG.\n\n(-1) times [ LOG (6) ]\n\n - LOG (6)\n\n\nTherefore, \n\n[ 1/4 LOG (16) ] - [ 2 LOG (6) ] + [ LOG (3) ] =     - LOG (6)
878117
depending on where your other bracket (]) is:\nlog[4,000(1.0042)^12t - 10,000(1.0063)^12t]= log1 is true\nwhen t=-164.08 or -36.6146
470727
log_7(7)+log_7(x)=1+log_7(x).
218601
Attached couple of links\n\nApart from these you can buy the book Primer on Logarithms (ISBN 8173714142)