what are logarithms and antilogarithms??
362293
The logarithm (see attached article 1) of a number x in base b is the number n such that b^n = x. It is usually written as log_b(x) = n (b in a subscript).\n\nFor example, log (100) = log_10 (100) = 2, because 10^2 = 100.\n\nThere is one special case - natural logarithm (in base e =~ 2.7182818). It has some properties, which are very useful in calculus. They are written as ln (x).\n\n\nAntilogarithm is a synonym of exponentiation or power.\nThis is an inverse function to logarithm.\nIf log_b(x) = y, then b^y=x.\nFor integer x, b^x mean multiplication b*b*b*...*b (x times).\nFor all real x, b^x are defined in the second attached article, part "Exponentiation with real exponents".\n\nHope it helps.
978128
log(base a) b = logb/loga.\n\nSo take the log of X and divide by the log of 1/3.  Then press the +/- button because you're being asked for the negative of that value.
521071
logarithm is used when u r dealing with exponential data, X^n.\n\nOne common use in financial mathematics, due to compounding, interest rates r defined in the form of force of interest, force=Log(1+i). Prices of shares r shown in log form often too.
941613
log_a b = x is the number to which you need to raise a to get b:\na^x =b\n...........
379984
Before you can compute a logarithm, you have to define a base that we can call b, where b > 0.  If the log base b of a number x is equal to k, then that means that x = b^k.  Since b is positive, then x has to be positive.\n\nIf a = sqrt(b), then that means that a*a = b.  If a is positive, then a*a is a positive times a positive which is positive.  If a is negative, then a*a is a negative times a negative which is positive.  Thus for the square root function to make sense, then b must be positive.
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.
1117027
Log0.78= -0.107905397309519598284728044078063\nLog2=0.301029995663981195213738894724493\nLog0.78/Log2= -0.358453970912476353890087260774269\n\nPut your cursor on the answer to see full length
812767
The logarithm is the mathematical operation that is the inverse of exponentiation, so you were not wrong.\n\nFor  y = log(base b)(x)\nThe inverse of a logarithm is called an antilogarithm or antilog; thus, x is the antilog of y in the base b. \ni.e x = b^y\n\nSimillarly in a prticular case of natural logarithm (base e)\n\nFor  y = log(base e)(x)\nx is the antilog of y in the base e. \ni.e x = e^y.\n\nI hope, I answere your question.
815091
First of all, we apply the power rule:  a*log(x) = log(x)^a\n\n     2*log(a(x+1)) = log(a(x+1))^2\n\nNext, we know that the sum of two log expressions can be written as a single product: log(a) + log(b) = log(a*b)\n\n     log(a(y+5)) + log(a(x+1))^2 = log[a(y+5)*(a(x+1))^2]\n\nYou may try some algebraic simplification, but I think that these concepts are the ones that are being tested for this problem.
756495
log(base x) y=[log(base 10)y]/[log(base10)x]. I hope you can take up from here!
443863
First off we need to remember what is meant by the logarithm of a number.  The logarithm of a number y in the base b is the exponent to which we have to raise b to in order to obtain y.  In other words the logarithm of y in the base b is the value of x in the equation\n\ny = b^x\n\nFor example what is the logarithm of 100 in the base 10?  Put y=100 and b=10 in the above equation - so we need to solve \n\n100 = 10^x\n\nyou should be able to see that x is 2 and so the log of 100 in base 10 is 2.  So if we wanted to find the logarithm of a negative number (say -1) in base 10 then we need to solve the equation \n\n-1 = 10^x\n\nIf we assume that x is a real number then this equation has no solution.  Thus the logarithm of negative numbers does not exist.\n\nHowever - when you study maths at a higher level you will eventually come across the idea of complex numbers.  If we allow x to be a complex number then we CAN take the logarithm of negative numbers and in base 10\n\nlog (-1) = 1.36437635 i\n\nwhere i is a special number known as the imaginary unit and is defined as \n\ni = sqrt(-1)\n\nSo in fact you CAN take the logarithm of negative numbers but only if you allow complex numbers in your solution.
783096
yes, it really is easier than you think:\na^(log_a(2/x) +log_a(x))=a^(log_a(2) -log_a(x) +log_a(x))\n  =a^(log_a(2))\n  =2\nnote that i used the property log(a/b)= log(a)- log(b) and a^log_a(x)=x
483658
It means that u have to solve for t....\n\n\n Is that log3(3t+1) means log (3t-4) base 3 or it means log3(3t-4) base 10?\n\n\n\nI assume  that log3(3t-4) is log (3t-4) base 3\nTo make the calculation easier i denoted log base 3 as lp\n\n       lp(3t-4) - lp(t+1) = lp 2\n       lp[ (3t-4)/(t+1) ] = lp 2     ->the rule of logarithm\n       Terminate the lp on both sides, we get\n\n                       (3t-4)/(t+1)= 2\n                              3t-4 = 2(t+1)\n                              3t-4 = 2t+2\n                             3t-2t = 2+4\n                                 t = 6
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.
781604
OK, there's 2 possibilities:\nIf you meant:\nlog (16^1/4), then you need to know the property of logarithms where:\nlog a^b = b log a.\n\nSo:\nlog 16^(1/4) = 1/4 log 16\nlog 10^(.01) = .01 log 10\nlog 2^(1/8) = 1/8 log 2\n\nIf you meant (log 16)^1/4, then I have no idea how to do it by hand.\n\nEdit:\nFurther, you can also do things like:\nlog 16 = log 2^4 = 4 log 2\nso 1/4 log 16 = 1/4 * 4 log 2 = log 2.
887169
You're getting into logarithm.  If X/D = R^Y, then Y = log(X/D), where log is the logarithm in base R.
659018
Yes, there is a formula to change the base of a logarithm, and here it is...\n\nIf you have an expression log(x) with base 3, the change of base formula allows you to re-express this in the form (lg x)/(lg 3), where lg refers to a logarithm with a base of 10.\n\nFor information on the change of base formula and other logarithmic rules, you may find this page useful:--\nhttp://en.wikipedia.org/wiki/Logarithmic_identities