how do you prove log(b^n) of X = 1/nlog(b) of x??
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
783096
yes, it really is easier than you think:\na^(log_a(2/x) +log_a(x))=a^(log_a(2) -log_a(x) +log_a(x))\n  =a^(log_a(2))\n  =2\nnote that i used the property log(a/b)= log(a)- log(b) and a^log_a(x)=x
477145
log(2x+1) - log5  = log x\nlog(2x+1) - log x = log 5\nlog((2x+1)/x)  = log 5\n(2x+1)/x  = 5\n2x+1 = 5x\n3x = 1\nx= 1/3
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)
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
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
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.
694278
Log(x + 2) + Log(x - 1) = Log(70)\nAs I explained before,\nLog (a0 + Log (b) = Log(ab)\nSo,\nLog(x + 2) + Log(x - 1) = Log [(x + 2)(x - 1)]\nLog [(x + 2)(x - 1)] = Log(70)\n(x+2)(x-1) = 70\nx^2 - x + 2x - 2 = 70\nx^2 + x = 72 \nx^2 + x = 72\nx^2 + x - 72 = 0\nx^2 - 8x + 9x - 72 = 0\nx(x - 8) + 9(x - 8) = 0\n(x + 9)(x - 8) = 0\nNow, either x + 9 = 0 or x - 8 = 0\nIf we put x + 9 = 0, we get x = -9\nIf we put x - 8 = 0, we get x = 8\nBut when we substitute x = -9 in the original problem we get Log (-7) + Log (-10) = Log(70)\nBut Log of negative numbers if undefined.\nSo, x = 8 is the only solution.
822813
Well the log is the inverse of an exponent right?\n\nLog base B (X) = N is equivilant to \nB^N = X\n\nYou want to find why Log 2 = .301 right?\nLet's turn it into an exponent.  I assume you're using log base 10 so we get:\n10^.301 = 2 (approximately) right?\nThis is because .301 is the exponent that you must bring 10 to the power of in order to get 2.  That's the best explanation I can offer.
368201
(1/2)logx = log(x^(1/2)) = log(sqrt(x))\n\nso for exemple 1/2log(9)=log(sqrt(9)) = log(3).\n\nsometimes it's useful, sometimes it's more useful to write the log of a sqrt as 1/2 the log of the argument.
750081
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
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.
458822
There's three log rules you need to know:\n\n1) ln a + ln b = ln (ab)\n\n2) ln a - ln b = ln (a/b)\n\n3) ln a^b = b×ln a\n   where "^" means "to the power of"\n\nWe can do this using the second rule backwards:\n\nln (1/2) = ln 1 - ln 2\n\nand ln 1 is zero.\n\nOr we can do it using the third rule\n[remembering that 2^(-1) = 1/2].  \nHence:\n\nln (1/2) = ln (2^-1) = -1×ln 2\n\nHope that helps!
978189
e^(ln5)=x   .  Now express this in log form by thinking of ln(5)\n                       as just an exponent.\n\n   lnx=ln5   so obviously\n\n   x=5.
881075
if these are decimal logarithms, i.e. log base 10, use the formula\n\nlog(ab) = log a + log b\n\nfor example,\n\nlog(417) = log (4.17 * 100) = log 4.17 +log100\n= .6201 + 2\n=2.6201\n\nlog0.417 = log 4.17 + log0.1 = .6201 -1 = -0.3799\n\nlog(.1) = log (10^(-1)) = -1
767881
ln( a^ b) = b ln a\n\nusing the above ln(y^52) = 52 ln y