science, digital computers are based on the binary numbering system which means that there are only 2 numbers available to the computer,0 and 1. when a computer scientist needs a logarithm, he/ she needs a log to base 2 which is not on any calculator. to find the log of a number to any base, we can use a conversion formula. log little b on bottum a= loga/ log b use this formula to find log2 10000
127485
log base 2 of 10,000 is log base 10 of 10,000 divided by log base 10 of 2, that's 4/0.3010 = (without a calculator) 13 or so.  13.29.
434596
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
491487
log(1) = 0 <1\nSo true for n = 1\n\nAssume true for n = k\nie logk < k\n\nThen (k + 1) = (k) + 1\n> log k + 1 \n= logk + log(base) \n= log(base*k)\n> log(k + 1) (base*k > k + 1 → base > (k + 1)/k = 1 + 1/k)\n\nTherefore if true for n = k then true for n = k + 1\n\nWell its true for n = 1 and therefore true for n = 2, 3, 4 .... all natural numbers n.
614173
Are you looking at computer code?  B indicates binary base and D indicates decimal.\nB100 = 1*2^2 + 0*2^1 + 0*2^0.\nD1000 is one-thousand.
652191
I suggest, without going into Decimal to Binary conversion, you use the calculator supplied with Windows.  If you pull up the calculator and then click on VIEW and select SCIENTIFIC, you'll get the conversion capabilities. Just type in the decimal number and then select binary. I also suggest you take the IP address and do it in sections, ie. the groups separated by the decimal point.  If you do the ENTIRE address, the BINARY result will be huge. This begs the question...WHY?\n\nGood luck.
404320
log 3^1/2 to the base 3\n=1/2 log 3 to the base 3\n=1/2
433445
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.
356335
There are infinite number systems simply because you can base a number system on any number except zero, but even 1 - the unitary system.  You've seen it - 1111 with a diagonal line through it for five.\n\nWe use the decimal number system, computers the binary, octal, and hexadecimal number system.\n\nNo matter which number system you use, the rules of math are the same.  For decimal you would carry ten, for binary, you'd carry 2, whatever the base of the system is.
184615
nope... 01 (in binary) = 1 (in decimal)\n\nand \n\n10 (on binary) = 2 (in decimal)
55752
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
476963
fist domain\n\nthe number we search the log must be positive so x>-2 \n\ndomain -2<x<infinity \nfor infinity log (base3), infinite is infinite. for x=-2 we tend to x+2 =0 and log (x+2 ) = -infinite  Vertical asypmtote for x=-2 \n\nthe equation can be rewritten  h(x)  = log (base 3) (x+2/7^3)
336011
get the n's on one side\nso\n17959/21558=(1.02/1.03)^n\nor\n.833055014=.990291262^n\nusing logs with change of base\nn=log (.833055014)/log(.990291262)\nor n=18.72
294923
Yes you can use any base logarithms to multiply and divide positive numbers.\n\nBy simplify, I presume you mean 'calculate the answer'.\n\n9.813617548 + 9.760194383 - (-7.398778275) = 26.97259021\n\ne^26.97259021 = 5.17663 x 10^11\n\n(I cheated - I used Excel)\n\nI guess you knew that to multiply, you add the logarithms and to divide, you subtract the logarithm.
145732
You can use logs.  By taking the log of any number, dividing it by, say, 2, then applying the result as an exponent, then you'll get a square root.  Try this first: 10^((log(2))/2) which will give you the square root of 2.\n\nAs for your answer:exp((ln(cell))/12)\nWhere exp() = exponent of e, e = 2.71...\nln() = natural log\ncell = cell reference in Excel
490095
2log(base3)x-log(base3) (x-4) = 2+log(base3)2\nlog_3 x^2/(x-4)=log_3 9+log_3 2=log_3 18\nx^2/(x-4)=18\nx^2=18x-72\nx^2-18x+72=0\n(x-12)(x-6)=0\nx=12\nx=6
542076
The logarithm equation you presented is the change of base law.\n\nC can be any number, commonly 10 or e  as our calculators only work in bases of 10 or e and this law helps to solve log equations of other bases. \n\ne is the natural logarithm number (its like the concept of pi, e= approx. 2.71....)
446501
In windows, open up the calculator.\nGo to scientific mode.\nSelect "Bin".\nType in the binary number.\nSelect "Dec"\nThe number is now displayed in decimal.\n\n4\n5\n12\n21\n15\n185\n154