maths help needed urgently..logarithms?
824045
i dont know where you got that formula from so i made up my own\n\n1200=400(2)^(x/3)\n\nI also dont know what you mean by m subscript 0 and K so im gonna skip those questions and move on to #3\n\n3=2^(x/3)\nlog3/log2=x/3\nx=4.75\n4.75*7=33days
140923
The other property of logarithms is that Log_x A - Log_x B = Log_x (A/B)\n\n\nso you can combine this rule with the other one i told you about in your other question...\n\nthe only thing is that u are working in reverse order\n\n\nLn 2y/x= Ln 2+ Ln y - Ln x\n\n\nThese are properties, they should be tought to you as principals, without them you cant solve logs:D hope this helps ;) if you think it does pls gimme 10 pts:P
7970
Log(a^b)= b*Log(a)\nLog(ab) = Log(a)*Log(b)\n\na) Log147= ?\n147= 3*7*7\n=> Log147= Log3 * (Log7)^2\n=> Log147= 2yw\n\nb) Log28= ?\n28= 2*2*7\n=> Log28= (Log2)^2 * Log7\n=> Log28= 2xw
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
765100
Have you studied logs yet?  You need to bring logs into this I think!\n\nRecall the log properties, as we will be using these:\nln(ac) = ln(a) + ln(c)\nln(a/c) = ln(a) = ln(c)\nln(a^r) = r*ln(a)\n\nSo you have y = -10*36^(1-2x) + 5\n\nFirst rewrite as this:  (5-y)/10 = 36^(1-2x)\n\nNow take the ln of both sides:\n\nln ((5-y)/10) = ln(36^(1-2x))\n\nNow use our log properties to get:\n\nln(5-y) - ln(10) = (1-2x) ln(36)\n\nDivide both sides by ln(36):\n\n( ln(5-y)-ln(10) ) / ln(36) = 1-2x\n\nHope this helped a bit.  I didn't solve everything for you, but hopefully this points you in the right direction.\n\nGood luck!
813587
2x=x(1.145)^t\nlog2+logx=logx+t log1.145\nt log 1.145t=log2\nt=log2/log1.145\n=0.3010/0.0588\n=5.2 years approx
51150
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
630919
all you have to do is use the change of base formula:\n\nlog5/log12=.6477
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
106841
Try the following resources:\n\n1. http://www.krysstal.com/logarithms.html\n2. http://www.physics.uoguelph.ca/tutorials/LOG/\n1. http://en.wikipedia.org/wiki/Logarithms
533385
Please believe TheoDorkx. He's the only one who understands logarithms.\n\nAnother way to solve logarithm problems is directly change from the logarithmic to the exponential form.\n\nln(2x) = 2 means that 2 is the power to which you raise the base of the logarithm, e, in order to get the result 2x.\n\nSo the equivalent exponential form is\n\n2x=e^2\n\ntherefore\n\nx=(1/2)e^2\n\nThis is the same thing TheoDorkx got with his approach. I offer it as a slightly different way to think about the problem.
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.
899135
1. Move the log part to the other side to make it positive, and move the 5 to the other side to get -2 = log(x+2)\nThe base of log is 10, so this means basically 10^-2 = x+2\n\n10^-2 is 1/100 or 0.01 = x + 2\n\n-1.99 = x\n\n2. 2e^(2x) + 5e^x - 3 = 0\n\nFactor by FOIL:  (2e^x - 1)(e^x + 3) = 0\n\nSo either e^x = 1/2 or e^x = -3 (which is impossible)\n\nYou can do ln (natural log) of both sides to "cancel out" the e:\n\nln(e^x) = ln (1/2) becomes x = ln(1/2)\n\n3. logx - 1/2log(x+4) = 1\n\nMinus becomes divide and the 1/2 becomes an exponent when combining logs.  So the left side simplifies to be:\n\nlog(x/(x+4)^(1/2))   Remember, 1/2 as an exponent means square root so this is\n\nlog(x/ sqrt(x+4)) = 1\n\nAgain, log has base 10 so this is really 10^1 = x/sqrt(x+4)\n\nSquare both sides: 100 = x^2/(x+4)\n\nCross multiply:  100(x+4) = x^2\n\n100x + 400 = x^2\n\nx^2 - 100x - 400 = 0  \n\nUse quadratic formula (be sure to eliminate negative answer(s) because you can't find the log of a negative number)
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)^
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)  )
829854
Use logarithms to bring exponents down where you can deal with them.  Since the natural log is base e, that's the way to go here.\n\ne^5x = 7\n\n5x = ln 7\n\nand solve from there.
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.