please help me to solve this Log (6 base) 16 if log(12base) 216=a?
360289
The correct answer is:   (12 / a) - 4\n\nThis can be checked by a = log 216/log 12 = 2.163\nlog 16/log 6 = 1.547\n12 / a - 4   = 1.547\n\nSolution method:\nNote that log(6)16 = log(6)2^4 = 4log(6)2\nSo if we can get log(6)2 in terms of "a" we can find our solution.\n\na =\n= log(12)216\n= log(12)6³\n= 3 log(12)6\n= 3 log(6)6 / log(6)12 ...........by change of base\n= 3 / log(6)12 ......................by log(6)6 = 1\n= 3 / (log(6)2 + log(6)6) ........by log ab = log a + log b\n= 3 / (log(6)2 + 1) ................by log(6)6 = 1\n\nRearrange \na = 3 / (log(6)2 + 1) \n\nto give\nlog(6)2 = 3 / a - 1\n\nThen\n\nlog(6)16 = 4 log(6)2 = 4 ( 3 / a - 1) = 12 / a - 4
880448
ln(x+6)-ln5=ln(x+6)/5  difference rule\nln(x+6)/5=ln(2x-7)  given\nsince the logs are equal removing the log sign\n(x+6)/5=2x-7\nmultiplying by 5\nx=6=5(2x-7)\nx+6=10x-35\nadding -6\nx=10x-41\nadding-x\n0=9x-41\nadding 41\n9x=41\ndividing by 9\nx=41/9
646100
log12.5base5 + log2base5 = log 25base5 = 5\n\nlog0.2base25 = log 5^-1 base 25 = -log5base25 = -.5\n\nlog27base3 + log9base3 -log54base? = 3 + 2 - ?\nyour question got cut off\n\nlog100base 10 = 2\n\nlog 1000 base 10 = 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.
805545
Part A\nlog(base3) 20 = (log 20) / (log 3)\nlog 20 = log (2*10) = log 2 + log 10 = 1.301\nlog 3 = 0.477\n1.301/4.77 = 2.727 --> about 2.73\n\nPart B\nlog(base 9) 100 = (ln 100) / (ln 9)\nln 100 = ln (10^2) = 2 (ln 10)\nln 9 = ln (3^2) = 2 (ln 3)\n(ln 10) / (ln 3) = 2.30 / 1.10 = about 2.10\n\nPart C\n1) log(base a) 6xy^5z^4 = log(base a) 6 + log(base a) x + log(base a) y^5 + log(base a)z^4 = log(base a) 6 + log(base a) x + 5 log(base a) y + 4 log(base a) z\n\n2) log (5a/4b^2) = log 5a - (log 4b^2) = log 5a - (log 4 + log b^2) = log 5a - log 4 - 2 log b\n\nPart D\n(1) 1/2 log a = log (a^(1/2))\nlog (a^(1/2)) - log 2 = log (a^(1/2) / 2)\n\n2) ln 2x + 3 ln x - 3 ln y = ln 2x + ln x^3 - ln y^3 = ln (2x*x^3)/y^3\n= ln (2x^4/y^3)\n\npardon me if you see typos.\nIt's 1:30 in the morning
824419
log 10 * log 10 = 1 because log (base 10) 10 = 1 \nand therefore 1 * 1 = 1\n\nlog 10 + log 10 = 1 + 1 = 2 \nand log 10 + log 10 = log (10*10) = log 100\nlog 100 = 2 means 10^2 = 100 noting again that log means log (base 10) meaning it is the power to which the base is raised to get its argument\n\n1000 = 10^3 so log 1000 = 3\n\nNow using the rules stated\n[ie  log a - log b = log (a/b)\nlog a + log b = log ab\nlog a^b = bloga]\n\nlog 3 + 3 log y = log3 + log (y^5) (since bloga = log a^b)\n= log(3 * y^5) (since log a + log b = log ab)\n= log (3y^5)\n\nThe log of a product = the SUM of the individual logarithms!\n\nlog(ab) = loga PLUS logb defintely NOT loga TIMES log b
889767
log2 8 is log8/log2\nlog2 32 is log32/log2\n\nnow just plug it in a calculator!\nyou should get 8.\n\nif you can't use a calculator, set it up like this:\n2^x=8 and 2^y=32\nthen add x and y (3 and 5)
1164623
s = c(1 - d)^n\n(1 - d)^n = (s/c)\nlog(1 - d)(s/c) = n\n\nANS : n = log(1 - d)(s/c) or n = log(1 - d)s - log(1 - d)c\n\nWhereas (1 - d) is the base\n\nor you can type it like this\n\nn = (log(s/c))/(log(1 - d))\nor \nn = (log(s) - log(c))/(log(1 - d))
707362
1)\nlog(base6)(40)= log(base10)(40)/log(base10)(6)\n\n2)  Factor 42 to  7*3*2.\n\n  ln(42)=ln(7*3*2)=ln7+ln3 + ln2\n\nFactor 45 to 3*3*5 .\n\n  log45=log(3*3*5)=2log3+log5
883916
Let log_2(x) denote the log of x with base 2.\n\nUse the identity that log_2(x) = ln(x) / ln(2). This identity holds because:\n\nRecall that, by definition, 2^(log_2(x)) = x.\n\nln(x) = ln(2^(log_2(x))) = (log_2(x)) (ln(2)), so dividing through by ln(2) shows the result.
245399
log base (1/5) of 78125 = -7\n\nlog|(1/5) 78125=-7
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Hiya, right here goes - \n\nyou want to find log10 . where log is to the base 2.\n\nfrom the prvious lecture notes you can see that the starting point fot the discrete logarithm is to fix a primative element which is dependent on which base you are working to. By definition this is an element g such that\n\ng^i ≠ 1 (mod p) for 1<i<p-2  ( and p>10 for it to work)\n\nin the example g=2 and p=11. p has been chose as 11 because\n\n 2^i ≠ 1 (mod p)  for 1<i<9\n\nhe point of the logarithm is then to find i such that 2^i ≡ 10 (mod p) trial and error gives that 2^5 ≡ 10(mod 11)\n\nso log10=5
962408
log4+xlogb=xloga\nx(loga-logb)=log4\nx=log4/(loga-logb)
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.
826204
Do you mean \n\nlog (base n) (x) = 1/(log (base x) (n))\n\nWell let log (base n) (x) = a\nTherefore x = n^a\n\nTake logs (base x)\nSo 1 = log (base x) (n^a)\n= a * log (base x) (n)\nBut a = log (base n) (x)\nTherefore log (base n) (x) * log (base x) (n) = 1\nie log (base n) (x) = 1/(log (base x) (n)) ............. QED
1173749
2\n\nwhen adding logs with similar base you multiply; 4*9 = 36\nlog (6) 36 = 2
898977
log(2x+1)-(1/3)log(x+1)\n=log(2x+1) - log(  (x+1)^(1/3)  )\n=log( (2x+1) /   (x+1)^(1/3)  )