Logarithms and Exponents?
362883
16^(3/4) means take a fourth root of 16, and then cube it.  Anytime you take an even root (square root, fourth root, etc.), you always have +/- answers.\n\nConsider that the square root of 4 is +/- 2.  That makes sense because if you square 2, it is 4, and if you square -2, it's 4 as well.\n\nNow consider the fourth root of 16.  That gives you +/- 2.  If you reverse it, you'll see it makes sense: 2^4 = 2*2*2*2 = 16, and (-2)^4 = (-2)(-2)(-2)(-2) = 16, since all the negatives end up cancelling out.\n\nSo for your original problem,\n\n(3x-1)^(4/3) = 16\n3x - 1 = 16^(3/4)\n3x - 1 = +/- 8\n3x = 1 +/- 8\nx = (1 +/- 8)/3\n= 3 or -7/3\n\nAnd if you plug in either value into your original equation, it works out.
449358
Hiya, right here goes - \n\nyou want to find log10 . where log is to the base 2.\n\nfrom the prvious lecture notes you can see that the starting point fot the discrete logarithm is to fix a primative element which is dependent on which base you are working to. By definition this is an element g such that\n\ng^i ≠ 1 (mod p) for 1<i<p-2  ( and p>10 for it to work)\n\nin the example g=2 and p=11. p has been chose as 11 because\n\n 2^i ≠ 1 (mod p)  for 1<i<9\n\nhe point of the logarithm is then to find i such that 2^i ≡ 10 (mod p) trial and error gives that 2^5 ≡ 10(mod 11)\n\nso log10=5
887169
You're getting into logarithm.  If X/D = R^Y, then Y = log(X/D), where log is the logarithm in base R.
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
486697
You do not need to use logs.\n27 can be written in base 3\n27 = 3^3\nTherefore,\n27^4 = (3^3)^4\nMultiply the exponents to get\n27^4 = 3^12\nThus, your equation is not\n3^x = 3^12\nWhenever the bases are the same, then the exponents are equal.\nTherefore,\nx = 12
494525
Think of it in terms of logarithms, since it is the inverse of exponents.\n\nStatement to prove:  1^infinity = 1\n\nRewritten in logarthmic form:  log[1](1) = infinity\n(Statement above is read: log base 1 of 1 equals infinity)\n\nWell that can't make sense, right?  It is a law of logs and of exponents that log[b](1) = 0 for any positive base b.  This is the same as saying b^0 = 1, for any b.  Also, log[b](b) = 1, for any positive b.  This makes since in b^1 = b.  Both of these laws would conflict this reasoning.  In fact, when b = 1, they conflict with one another.  Therefore, lets look at it with a variable instead.\n\nProve:  1^infinity = X\n\nRewrite in logartithmic form: log[1](X) = infinity\n\nNow, what value of X will give you infinity?\nCan you determine that?\n\nI would say it is indeterminate :)
303836
Nope,\n\n   The log function blows up to infinity.  (If you don't believe this, give me a value for y=f(x) = K * log(x).  Ha ha! I can give you a value for x such that f(x) is greater than the value you gave me! )\n\n  Your pal from Calc I\n   cuntsucker
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
1187309
I think I did it.  If y=1/log(x), then a few manipulations gave me\n\nx=10^(1/y).\n\nI'm assuming by log you mean log to the base 10.  I suppose mathematicians will start fussing about what happens at y=0, but a physicist like me doesn't worry that much about infinities ...
693908
Logs are just equivalent to power statements. \nlog_a (b) = c means a^c = b.\nSo that equation just means x^2 = 2x-3.\nNow thats just a quadratic;\nx^2 - 2x + 3 = 0\nHowever that has no real roots. So there are no such values of x. But hopefully you get the idea.
437664
The secret is to understand some of the rules of logarithms and adjust translate the numbers to make your lob easier.\n\nThe rules:\nlog(a*b) = log(a) + log(b)\nlog(a/b) = log(a)- log(b)\nlog(a^n)=nlog(a)\n\nIf I understand your problem:\n\nx= [(86400)*sqrt(365.25)]/[ (.079)^11]\n\nso log(x) = log [.....]\n\n     = log(86400)+log(356.25^.5) - log(0.79^11)\n     = log(86400)+.5*log(356.25) - 11*log(0.79)\n\nuse log tables to get the logs and add them up to get the log of x.  Use the tables again to find x.\n\nYou can make other approximations, i.e., 356.25 is close to 19^2 (=361)\nand 86400 is close to 300^2 (=90000) and continue from there.  \n\nHope that helps.
292874
exponents. for example 2^2 would be 2 to the power of 2 which would be 4. 2^3 would be 8. I hope this helped.
987583
2^(x+1) = 5^(1-2x)\n\nlog 2^(x+1) = log 5^(1-2x)\n\n(x+1) log 2 = (1-2x) log 5\n\n(log 2 + 2 log 5)x = log 5 - log 2\n\nx = (log 5 - log 2)/(log 2 + 2 log 5)
836453
a^24b^24\njust multiply the exponents\n\nhope thie helps!!!
860995
Do the log of both sides.  The exponent will become multiplied by the log of the base.\n\nlog(5/6) = -t/5760 log 2\n\nDivide both sides by log 2, then multiply both sides by -5760 to get t.
364460
log(x) in base a (for a>0) is simply ln(x)/ln(a). So if you are able to calculate natural logarithms, you can calculate logarithms in any base.\nA very simple technique to calculate natural logarithms is to use the identities:\nd/dx(ln(x))=1/x and ln(1) = 0. This implies that to calculate ln(y) for a given y>0, you can use finite approximations to the integral:\n\nintegral form 1 to y of 1/x = ln(y)\n\na very simple approximation is the following: for a given integer (big!) k of your choice, \n\nln(y) = sum for i=0 to k-1 of k/(k-i(y-1))\n\nHope this helps
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)