If log xa  = 3 then the value of log x1/a  is?
470800
logx 1/a = (logx 1) - (logx a)\n= 0 - 3\n= -3
1027143
log 2 x^2+1-log2^x\nlog 2 + log X^ + log 1 - (log 2^ + log X)\n=log 2 + A log X + log 1 - A log 2 - log X\n=log 2 + log 1- log X + A log (X - 2)\n=log (2 x 1) - log x + A log (x -2)\n=log 2/x + A log (x-2)\n=log 2/x + log (x-2)^\n= log (2/x X (x-2)^) = log 2/x(x-2)^
439071
a) log3(x-24) is valid for all x but x = 24\nb) log3(x) is valid for all x but x = 0\nc) x-24  = 3^4 solve for x :)\n\nRival
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
962408
log4+xlogb=xloga\nx(loga-logb)=log4\nx=log4/(loga-logb)
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)
771001
In the first, 1/e = e^-1, and (e^-1)^9 = e^-9.  Since ln x is the inverse of e^x, ln(e^-9) = -9\n\nIn the 2nd, .01 = 10^-2, so log[ (10^-2)^(1/3)]= log 10^(-2/3) = -2/3
780572
First let's remember some rules of logs.\nThe log of product is the sum of the logs.  That is,\nlob(a*b) = log(a) + log(b)\nThe log of a quotitent is the difference of them.  That is, \nlog(a/b) = log(a) - log(b)\nAnd finally, if you have an exponent, you can bring it in front of the log.\nlog(a^n) = n*log(a)\nRemember that you can only take the log of positive numbers.\nln[ x^2 / [sqrt(x) (2+x)^2] ]\n= ln(x^2) - ln[ sqrt(x) (2+x)^2]\n= 2ln(x) - [ln(sqrt(x)) + ln(2+x)^2]\n= 2ln(x) - (1/2)ln(x) - 2ln(2+x)\nYou can combine 2ln(x) - (1/2)ln(x)\n2ln(x) - (1/2)ln(x) = (2 - 1/2)ln(x) = (3/2)ln(x)\nAnswer: (3/2)ln(x) - 2ln(2+x)\nNote: sqrt(x) = x^(1/2)
808711
Welcome!!\n\nHere is the solution:\n\nlog(x+1) = log3 - log(2x-1)\nlog(x+1) + log(2x-1) = log3\nlog[(x+1)(2x-1)] = log 3\n\ntherefore,\n\n(x+1)(2x-1) = 3\n2x^2 - x + 2x -1 = 3\n2x^2 + x - 4 = 0\n\nyou need to use the quadratic formula to solve this:\n\nx = {-1 ± sqrt[1 - 4(2)(-4)] } / 2*2\nx =[ -1 ± sqrt(33)] / 4
808044
Take the log of both sides:\nlog(0.01) = log(2.2^-t/8)\n\nRemember that log(a^b) = b * log(a), so bring the -t/8 in front:\nlog(0.01) = -t/8 * log(2.2)\n\nNow just solve algebraically:\n-t/8 = log(0.01) / log(2.2)\nt = -8 * log(0.01) / log(2.2)\n\nlog(0.01) = -2\nlog(2.2) ≈ 0.342422681\n\nt = (-8 * -2) / 0.0.342422681\nt = 16 / 0.342422681\n\nTherefore:\nt ≈ 46.7258768\n\nAs a double check, you can plug it in your calculator:\n2.2^(-46.7258768/8) = 0.01
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.
575720
take logbase6 of 135 = x\n\nlog 135/log 6 = x\n2.737 = x\n\nThen substitute this into original to get:\n2^(log135/log6) + 1.3^(log135/log6)-2 \n\nIt should prove that it equals 30, which proves that both statements are ture.  However, I am getting that it is not equal to 30...so I think the proof is impossible,  unless your formula is typed incorrectly.
726536
log 3x² - log 9x = 2\n\nSo log (3x²/9x) = log 100 (Since log a - log b = log (a/b) and 100 = 10²)\nThus\n3x²/9x\n= x/3 \n= 100\nSo x = 300
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.
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
1060074
d (ln(x)) / dx = 1/x\n\nThe derivative of the natural log of x with respect to x equals 1 / 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