If log a2 = A, log a3 = B and log a5 = C then log a0.006 expressed in terms of A, B and C is?
412613
B-2A-3C = log a(3/(2^2*5^3) = loga0.006\n\nso, answer is D
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
641969
log 35 = log (7 * 5) = log 7 + log 5\n\nlog 8 = log (2^3) = 3 log 2 = 3 * 0.3010\n\nlog 5 is already given isn't it. \n\nand by the way you have written the value of log 7 wrongly. the value that you have given is of log 3.
545001
1. 6^(-.2x)=10\n(-.2x)ln(6)=ln(10)\nx=-5*ln(6)/ln(10)\n\n2. 7log_5(x)=18\nlog_5(x)=18/7\nx=5^(18/7)
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!
916994
(log x)^3 = log (x^9)\n(log x)^3 = 9 log x\n(log x)^2 = 9\nlog x = +/- 3\nx = 10^3 = 1000  or x = 10^-3 = .001
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)  )
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
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)
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
760330
First let me assert that \n  log(base b)(a)=1/log(base a)(b) as follows:\n\n let log(base a)b=z\n   a^z=b\n   a=b^(1/z)  . Now sub back for z giving\n   a=b^(1/log(base a)(b))  and convert to log form of base b\n 1)   log(base b)(a)=1/log(base a)(b)  \n\n Now follow this: \n    let  log(base c)(x)=y\n        c^y=x   Now take log(base a) of both sides\n    ylog(base a)c=log(base a)x  and  sub back for y\n log(base a)x  = log(base c)x{log(base a)c} and\n use 1) to convert log(base a)c giving\n   2)  log(base a)x  = log(base c)x/{log(base c)a.\n\n     Lastly, into 2)  substitute X^3+1 for x,  4 for a and 10 for c\n\nand we have \n  log(base 4)(x^3+1)  = log(base 10)(x^3+1)/{log(base 10)4}
703263
log a (x)= 3.58 and log a (y)= 4.79,\nloga(y/x)=loga(y)-loga(x)\nloga(y/x)=4.79-3.58\nloga(y/x)=1.21
850657
In the logarithm, there are several number of rules you have to remember.  One of them is by combining two or more logarithms into one.\n\nFor example,\nlog (a) b + log (a) c = log (a) bc\nlog (a) b - log (a) c = log (a) (b/c)\na is the base of the logarithm.\nb and c is any value more than 0.\nThis kind of combination can only be done if and only if the logarithms have the same base.\n\nThe other rule involved in solving your question is as following:\nc log (a) b = log (a) b^c\na is the base of the logarithm.\nb is any value more than 0.\nc is any real number.\n\nIn your case,\nlog (a) (y + 5) + 2 log (a) (x + 1)\n= log (a) (y + 5) + log (a) (x + 1)^2\n= log (a) (y + 5)(x + 1)^2\n\nIf you would like to multiply it, \nlog (a) (y + 5)(x + 1)^2\n= log (a) (y + 5)(x^2 + 2x + 1)\n= log (a) (x^2y + 2xy + y + 5x^2 + 10x +5)\n= log (a) (x^2y + 5x^2 + 2xy + y + 10x +5)\n\nHowever, this only further complicate the answer, therefore log (a) (y + 5)(x + 1)^2 would be the best answer.
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
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.
911082
logx+log(x+48)=2\n=>log{x(x+48)}=2\n=>log(x^2+48x)=2\n=>x^2+48x=10^2\n=>x^2+48x-100=0\n=>x^2+50x-2x-100=0\n=>x(x+50)-2(x+50)=0\n=>(x+50)(x-2)=0\nhence,\nx=-50 or x=2
118410
...=(logA+logA)-(logB-logB-logB)+1/5logC=(logA^2-logB^3)+1/5logC\n= log(A^3/B^3)+1/5logC =log(A^3/B^3)+logC^1/5=log[(A^3*C^1/5)/B^3]\nNow if you know to transform C^1/5 (i don't really remember now)it becomes even simpler.