logarithmic question on how to enter it into my calculator, its college algebra?
968058
Type 1.16 log / 4 / 1.0125 log =\nThis is 3 approximately (2.9869...)\n\nTh.
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
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.
790275
log(9)3\n=\nlog(9)9^1/2\n=\n1/2*log(9)9\n=\n1/2\n[as log(m)m=1]
593752
log ( sqrt (x+ 1) *x^5)
218601
Attached couple of links\n\nApart from these you can buy the book Primer on Logarithms (ISBN 8173714142)
921542
Logarithms are exponents, so they follow the rules of exponents.  For instance,\n\n(x^a)*(x^b) = x^(a + b)\n\nTherefore, \n\nlog a + log b = log(ab)\n\nAlso, every exponential equation has a logarithmic form, and vice versa.  So, assuming we're talking base 10,\n\nlog x = y\n\nmeans the same as\n\n10^y = x\n\nSo,\n\nlog a + log b = log (ab)\n\nIn your case:\n\nlog(x) + log(x + 15) = 2\n\nlog(x(x+15)) = 2\n\nIn exponential form, this becomes\n\nx(x + 15) = 10^2\n\nx^2 + 15x = 100\n\nx^2 + 15x - 100 = 0\n\n(x + 20)(x - 5) = 0\n\nx = -20 or x = 5\n\nSince you can't take the log of a negative number, we reject x = -20, so the answer is\n\nx = 5.\n\nDoes that help?
178809
Have you considered some sort of logarithmic compression?  for example, 10*log10(x+1)\nthen 10*log10(1+1) = 3\n10*log10(3+1) = 4\n10*log10(2+1) = 3.7\n10*log10(2.5+1) = 3.?\n10*log10(400 +1) = 26\n10*log10(500 +1) = 27\n\nother types are possible using the natural instead of the base 10 log.   there may be some rounding errors in my numbers, i wasn't using a calculator.\n\nI hope I didn't misunderstand your qustion.
808711
Welcome!!\n\nHere is the solution:\n\nlog(x+1) = log3 - log(2x-1)\nlog(x+1) + log(2x-1) = log3\nlog[(x+1)(2x-1)] = log 3\n\ntherefore,\n\n(x+1)(2x-1) = 3\n2x^2 - x + 2x -1 = 3\n2x^2 + x - 4 = 0\n\nyou need to use the quadratic formula to solve this:\n\nx = {-1 ± sqrt[1 - 4(2)(-4)] } / 2*2\nx =[ -1 ± sqrt(33)] / 4
878117
depending on where your other bracket (]) is:\nlog[4,000(1.0042)^12t - 10,000(1.0063)^12t]= log1 is true\nwhen t=-164.08 or -36.6146
982509
Let log2 5 = x\n\nThen 2^x must be 5\n\nTake log of both sides but use log10 (common log, on calculator)\n\nx log 2 = log 5\n\nDivide both sides by log 2
812579
Adding logs is the same as taking the log of the two numbers multiplied. \nSubtracting the logs is the same as taking the log of the ratio (left number divided by the right number).\nlog (a) 15 - the log (a) 3 = log (a) 15/3 = log (a) 5.
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.
31425
log-1(z) is just b^z where b is the base of the logrithm system you are using.  If you are using "Common" logs, its 10.\n\n\nHow you compute it is going to vary with your calculator\n\nOn the caluclator that comes with Windows:\nSelect Scientific View.\nEnter 10\nClick on the x^y key\nEnter z (the value you want to take log-1 of)\nPress equal.\n\nTo test it, pick a number, say 5, enter it\nClick on the log key. Result will be .69897000...\nSave this for future use by clicking on M+ (add it to your saved memory)\nClear.\nEnter 10.\nClick x^y.\nClick MR (Memory recall) \nClick =. Result will be 5.
807860
Yes, though don't forget the y on the second log:\n\nlogb x^2/3 + logb y^3/4\n\nAlso, if you would have had:\nlogb x^2/3 / y^3/4 it would equal logb x^2/3 - logb y^3/4\n\nlog rules are just like exponent rules: multiplication = addition and division = subtraction
114069
log(a*b) = log(a)+log(b)\n\nlog(a/b) = log(a)-log(b)\n\nSo these work when expanding (left to right) and condensing (right to left).\n\n\nThanks for the update: So using your example:\n\n2ln3 - ln5 = ln(3^2) - ln5 = ln9 - ln 5 = ln(9/5)\n\nThe reason i converted 2ln3 to ln9 is because the condensing form only works when the coefficients of the logs are identical (ie 1 in this case). The formula for doing this is N*ln(x) = ln(x^N)\n\nNext, \n\nln(7X/3) = ln7 + lnX - ln3 (or ln(7/3) + lnX)\n\nAs you can see, the above form at 3 factors inside it - the expansion form will work for however many factors you want to use.\n\nIf you plug in the numbers into a calculator then you will see that this works.
958049
[Log(11)3 - 7]/0.35 = x\n\nRead: Log base 11 of 3 minus 7 all divided by 0.35 equals x