how do you work out 1/2 a log of any given number?
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.
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
574952
Use your logorithm rules to combine the terms on the lefthand side into one term\n\nlog (x^2/(8x)) = log (x-10)\n\nNow, if the logorithms are equal, the arguments must be equal\n\nx^2/(8x) = x-10\n\nCross-multiply\n\nx^2 = 8x^2-80x\n\n7x^2-80x=0\n\nx(7x-80)=0\n\nSolving this, you get x=0 or x=80/7. but you have to reject x=0 because it's not in the domain of the original expression (you can't have log(0)).\n\nso x=80/7
994994
If you are saying \n\n1.) (log(.78))/(log(2)) = -.35845397... or log(2)(.78)\n2.) (log(.80))/(log(2)) = -.32192809... or log(2)(.80)\n\nbut if you mean\n\n1.) (log(78))/(log(2)) = 6.2854022... or log(2)(78)\n2.) (log(80))/(log(2)) = 6.32192809... or log(2)(80)
47515
There are lots of ways to find logarithms without a calculator. After all, people had to do this when there were no calculators. \n\n1. Logarithm tables are in books and all over the place. Here is an example of a log table on the internet:\n\nhttp://www.sosmath.com/tables/logtable/logtable.html\n\n2. My advisor in graduate school actually memorized a lot of values from the log tables so he could do very fast mental calculations. \n\n3. Slide rules are easy ways to get approximate logarithm values. \n\n4. There are numerical methods for approximating logarithms that you can track down, for various values of the argument. For example, look at \n\nhttp://www.maths.abdn.ac.uk/~igc/tch/ma1002/appl/node59.html\n\nhttp://mathworld.wolfram.com/NaturalLogarithm.html\n\nhttp://www.informatik.tu-darmstadt.de/ftp/pub/TI/reports/MarkusMaurer.diss.pdf\n\nAfter all, the calculator has to use some method to produce or approximate the logarithm, and all one has to do is to use similar methods to find the logarithm by hand, as long as they do not require an excessive amount of computing.
815275
Using the quotient rule,\n\nlog(2x-1) - log(x-3)=1\nlog((2x-1)/(x-3)) = 1\n\nThen change to exponential form:\n\n10^1 = (2x-1)/(x-3)\n\nthen solve:\n\n10(x-3)=2x-1\n10x-30=2x-1\n8x=29\nx=29/8
646100
log12.5base5 + log2base5 = log 25base5 = 5\n\nlog0.2base25 = log 5^-1 base 25 = -log5base25 = -.5\n\nlog27base3 + log9base3 -log54base? = 3 + 2 - ?\nyour question got cut off\n\nlog100base 10 = 2\n\nlog 1000 base 10 = 3
742533
Do you know your exponent rules? The same apply for logs. Multiplication means addition (this is why x^5 times x^2 is x^7 and NOT x^10) and division means subtraction (this is why x^16 divided by x^4 is x^12 NOT x^4).\n\nKeeping these rules in mind, I'll show you some examples of how these rules can be applied.\nlog x + log y = logxy\nlog 2x + log 4x = log8x^2\nlog12 = log6 + log2\nlog 4x - log 9x = log4x/9x\nlog2 - log4 = log2/4\nlog10x/3x = log10x - log 3x\n\n\nYou can also pull out numbers to the front of the logs like so:\nlog10x = 2log5x\nRemember with this rule: you can only pull NUMBERs out, NOT variables!\nSome more examples of this:\n4log4x = log16x\nlog45x = 9log5x\n2log6x = log12x\n\n\nNow I'll give you an example of all of them together:\n2log5x + 2log6x\nlog10x + log12x\nlog120x^2\n\nAnother example:\n4log4x - 2log2x\nlog16x - log4x\nlog16x/4x
82029
it's a simple work: I'll give you a suggestion, but you should try to do it by yourself.\n\ny intercept: you find it by setting x=0 so you get\n\ny= log(base2)(2) + 3\n\nThe logarithm is simple to calculate ---> get an integer answer.\n\n---\n\nx intercept: you must set y=0 this time.\nYou get the logarithmic equation\nlog(base2)(x+2) = 3 ---> to solve for x, let each member be the exponent of a power with base 2:\n\n2^(logbase2(x+2)) = 2^3\n\n---> use definition of logarithm (why did we choose a power with base 2?) and get to an integer solution.\n\nGood work.
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
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)  )
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
550920
5^x=.125\nTaking log (base e) on both sides\n\n-> xlog5=log(0.125)\n-> x= log(0.125)/log5 = -1.291099302
968058
Type 1.16 log / 4 / 1.0125 log =\nThis is 3 approximately (2.9869...)\n\nTh.
968865
10 = -20 log x\nlog x = 10/-20 = -2\n10^(log x) = 10^(-2)\nx = 10^(-2) = 1/10^2 = 1/100
586612
remember your natural logs\n625^x = 1/125\nxln625 = ln(1/125)\nx = ln(1/125)/ln625 = -3/4
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