simplify logarithmic expression?
820123
log 10 = 1\nln e^-2 = -2\nlog(sub 11) 1 = 0\n\n... -1
627488
you can change the base...\nln x = (log2 (x))(ln 2)
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.
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.
462623
>> log (x+3) + log (2-x) = 1\n\nlog( a b ) = log( a ) + log ( b ) so\nlog (x + 3) + log (2 - x) = log( (x + 3)(2 - x) ) = 1\n\nlog(e) = 1 if using natural logarithms, log(10) = 1 if using base 10 logarithms\n\nFor the former you have (x + 3)(2 - x) = -(x^2) - x + 6 = e and for the latter you have -(x^2) - x + 6 = 10.  These are both quadratic equations which should be easy to solve.  For natural logarithms you have\n\n-(x^2) - x + 6 = e\nx^2 + x = 6 - e\nx^2 + x + (1/4) = (6 + (1/4) - e) = (25/4) - e\nx^2 + x + (1/4) = (x + (1/2))^2 = (25/4) - e\nx + (1/2) = +/- sqrt( (25/4) - e )\nx = -(1/2) +/- sqrt( (25/4) - e )\n\nAs a final check on suitability of these values, we see that numerically x = 1.379... or x = -2.379... and so both x + 3 and 2 - x are positive for both values of x and so you have no problem taking log( x + 3 ) and log (2 - x) with either value for x.\n\nWith base 10 logarithms, completing the square [i.e., solving the quadratic] would lead to complex roots:\n\n-(x^2) - x + 6 = 10\nx^2 + x = -4\nx^2 + x + (1/4) = -4 + (1/4) = -(15/4)\nx^2 + x + (1/4) = (x + (1/2))^2 = -(15/4)\nx + (1/2) = +/- (1/2)sqrt(15) i\nx = -(1/2) +/- (1/2)sqrt(15) i\n\nSo probably the natural logarithm was intended. :-)\n\n>> 2^x+1 =5^1-2x\n\nIf you mean (2^x) + 1 = (5^1) - 2x then you have\n(2^x) + 1 = 5 - 2x\n(2^x) = 4 - 2x\nand "by inspection" x = 1 is a solution.\n\nIf you graph both y = 2^x and y = 4 - 2x you find only one point of intersection; alternatively, the derivative of 2^x - 4 + 2x is strictly positive, so the function is strictly increasing and continuous and so has a single zero.\n\nEither way x = 1 is the only real solution to\n(2^x) + 1 = (5^1) - 2x\n\nHowever, if you meant\n2^(x + 1) = 5^(1 - 2x)\nthen you get (x + 1) log(2) = (1 - 2x)log(5)\n(log(2) x) + log(2) = log(5) - (2 log(5) x)\n(2 log(5) + log(2)) x = log(5) - log(2)\n(log(5 * 5 * 2)) x = log(5 / 2)\nx log(50) = log(5/2)\nx = log(5/2) / log(50)\n\n>> p = a/1+be^-kt \n\nIf you meant p = a / (1 + b(e^(-kt)))\nthen with all log's to base e you have\n\na = p (1 + b(e^(-kt)))\nb = ((a / p) - 1) (e^(kt))\ne = inverse log of ( log(b) - log((a / p) - 1) ) / (kt)\nk = - (1/t) log(((a / p) - 1) / b)\np = a / (1 + b(e^(-kt)))\nt = - (1/k) log(((a / p) - 1) / b)
586612
remember your natural logs\n625^x = 1/125\nxln625 = ln(1/125)\nx = ln(1/125)/ln625 = -3/4
795919
log a (7x+1)=log a (4x+16)\n7x+1= 4x+16\n3x = 15\nx= 5\n`
956834
For log(x+8) to be log x +log 8 ,x+8 must be equal to 8x\nWe can write it algebrically as log(x+8)=logx+log8=log(8x)\n=>8x=x+8\n=>7x=8    or,x=8/7
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]
179378
All you do is take the natural logarithm (ln) of both sides.\nBecause the ln of e to any power is just that power it becomes this:\nln (e^-x) = ln(6) \nThe ln sort of "cancels out" the e, leaving you with the power so...\n-x = ln(6)\nMultiple both sides by negative 1 and you're done, unless you need to approximate ln(6).
364460
log(x) in base a (for a>0) is simply ln(x)/ln(a). So if you are able to calculate natural logarithms, you can calculate logarithms in any base.\nA very simple technique to calculate natural logarithms is to use the identities:\nd/dx(ln(x))=1/x and ln(1) = 0. This implies that to calculate ln(y) for a given y>0, you can use finite approximations to the integral:\n\nintegral form 1 to y of 1/x = ln(y)\n\na very simple approximation is the following: for a given integer (big!) k of your choice, \n\nln(y) = sum for i=0 to k-1 of k/(k-i(y-1))\n\nHope this helps
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
894472
Use a calculator.  Enter [2] [log] for the first one.
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.
574952
Use your logorithm rules to combine the terms on the lefthand side into one term\n\nlog (x^2/(8x)) = log (x-10)\n\nNow, if the logorithms are equal, the arguments must be equal\n\nx^2/(8x) = x-10\n\nCross-multiply\n\nx^2 = 8x^2-80x\n\n7x^2-80x=0\n\nx(7x-80)=0\n\nSolving this, you get x=0 or x=80/7. but you have to reject x=0 because it's not in the domain of the original expression (you can't have log(0)).\n\nso x=80/7
746085
taking log on both sides\nlny=(x-5)ln2\n1/y(dy/dx)=ln2\ndy/dx=yln2=2^(x-5)ln2
754397
ln(x+3) - ln(x) = ln[(x+3)/3]\n\n2ln(2) = ln(2²) = ln4\n\nSo, if ln[(x+3)/x)] = ln4, then\n\n(x+3)/x = 4, or\n\nx+3 = 4x, or \n\nx = 3x, or x = 1\n\nI used the properties of logs to help me solve this.  I think you tried to make it too hard on yourself (which is very easy to do when dealing with the hated logarithms!).  Also, you brought e into it when you didn't really need to.  Since we had ln(x) = ln(y), essentially, then from there we can conclude that x=y without bringing e into play.  Anyhoo, here's the log properties I used in this solution.  (I use the word log but it applies to ln as well.):\n\nlog(a) - log(b) = log(a/b)\n\nrlog(a) = log(a^r)\n\nHope that helped!\n\n~ ♥ ~