Answer my query on logarithm?
968809
The logarithm equation you presented is the change of base law.\n\nC can be any number, commonly 10 or e  as our calculators only work in bases of 10 or e and this law helps to solve log equations of other bases. \n\ne is the natural logarithm number (its like the concept of pi, e= approx. 2.71....)
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)  )
574891
im pretty sure that you will have to use the parts method to solve it. but the answer is\n\nx(log(1-ie^(ix))-log(1+ie^(ix))+i(Li(-ie^(ix)-Li(ie^(ix))\n\ni=imaginary unit\nlog=natural logarithm\nLi=polylogarithm\n\nyou can check my answer to see if its right because i also thought that it was wrong. i dont know how the imaginary units could have shown up but o well.
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
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.
147502
log x goes to negative infinity\nbut with the squared it becomes positive.\nSo the limit is +infinity
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
584413
log(x-4) - log(3x-10) = log(1/x)\nlog((x-4)/(3x-10)) = log(1/x)\n(x-4)/(3x-10) = 1/x\nx(x-4) = 3x-10\nx^2 - 4x - 3x + 10 = 0\nx^2 - 7x + 10 = 0\n(x-2)(x-5) = 0\n\nx = 2 or x = 5.\n\nedit: Sorry, ervin is right.  I thought I had checked, but apparently I wasn't careful enough.  The only solution is x=5 since you cannot take the log of negative numbers.
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.
499879
I'm trying to guess what the problems you were asked to do were, because the information you gave in posing the question was incomplete.\n\nIf f(x) = log(x)  (for log of any base)\n\nthen\n\nf(1/b) = log(1/b) = log(b^(-1))=-log(b)=-f(b)\n\n\nlog(MN^2) = log(M) + 2 log(N)\n\nor\n\nlog((MN)^2) = log(M^2) + log (N^2) = 2log(M) + 2log(N)\n\nThese are applications of the laws of logarithms.
437222
Rules of logs:\n\n--log (a*b) = log a + log b\n--log (a/b) = log a - log b\n--log (a^b) = b log a\n\nSo log (y^2 / x^3) would break down as\n\nlog (y^2 / x^3) = log y^2 - log x^3\n=2 log y - 3 log x\n\nwhich is answer C in your list.
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.
148470
To quote one of my old teachers, who made a lasting impression by screaming the following at our class:\n\nLOGRITHMS ARE EXPONENTS!!!!!!\n\nIf you have an exponent, then you are taking some number to a power.  That number is known as the base.\n\nYou can use any positive number for the base but usually log refers to log base 10. ln is always base e, where e = 2.7...\n\nThat is, saying ln(x) = y is the same thing as saying x = e^y. (y is a logrithm so y is an exponent!)\n\nHistory:\nNatural logs originate in the search for an integral for 1/x. \nIf you define a function, F(x), such that F'(x) = 1/x and start looking at its properties, you find that F(ab) = F(a) + F(b) and F(a^n) = nF(a). The behavior you would expect from a logrithm function. When you calculate the base for that function, you get e.
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))
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
338108
A Natural log. It is expressed as: ln()\nFor example: ln(e^2) = 2
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