why log cannot take -ve values and o also?
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.
41347
The trick to remembering the log rules is that they are the same as the exponent rules since logs are just rewritten exponents.\n\n(x^2)(x^3) = x^(2+3) = x^5\nlog [(a)(b)] = log a + log b\n\n(x^4)/(x^2) = x^(4-2) = x^2\nlog [a/b] = log a - log b\n\n(x^3)^2 = x^[(3)(2)] = x^6\nlog[a^b] = b[log a]\n\nThe rules for log and ln are the same since the only difference is log has base 10 (unless otherwise noted) and ln has a base of e.\nThere are also the two inverse functions with log and ln
881075
if these are decimal logarithms, i.e. log base 10, use the formula\n\nlog(ab) = log a + log b\n\nfor example,\n\nlog(417) = log (4.17 * 100) = log 4.17 +log100\n= .6201 + 2\n=2.6201\n\nlog0.417 = log 4.17 + log0.1 = .6201 -1 = -0.3799\n\nlog(.1) = log (10^(-1)) = -1
458677
There's a very explicit algorithm which explains\nhow to handle number, operators (including left\nor right precedence), functions, etc. on the \nWikipedia site for RPN (see reference #1).\n\nMostly, it's very straight-forward.  If it's an operand\n(ie. a number), then push it on the stack.  If it's an\noperator, then pop off as many operands are needed and\nevaluate.\n\nBut do look at the full algorithm for the details.\n\nOh yeah, for a full-blown implementation, see the source\ncode to the Unix "dc" (desk calculator) program.  The\nSolaris source is at reference #2.
424839
Looks like it suppose to print properties for a file with the name  you pass as the first arguement\n\nIt appears to be depended on what ls -l <filename> outputs and may not work on all version of Unix.
790661
Use 'Save As' and then choose 'JPEG' as the saved file format.\nEasy as falling off a log.
608949
cant answer secifically, but you should be able to back calculte taking elements present today and using the half lives as your main variable.
44336
One solutions is:\n\nGet the formula from the user then parse it. You have to write a parser.\n\nAt run time, todays languages won't be able to get forumula and input from those variable.\n\nI am a Software Engineer and I know you can't do it with C, C++, Java, any .Net...\n\nGood Luck
902324
Hello dear \n\n▪log ( x +3 ) + log (x) = 1 \n{ as you know if base is a so log (xy) = log x + log y }\n▪log x(x + 3) = log 10 \ncancel " log " \n▪x(x+3) = 10 \n▪x ² + 3x - 10 = 0\n▪(x + 5)(x - 2) = 0\n▪x = + 2 \n▪x = - 5  { cant accept it ,  because 0 < a = b   <=>  log a = log b }\n\nGood Luck.
894472
Use a calculator.  Enter [2] [log] for the first one.
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
890606
=====\nThis behavior occurs because the operating system does not allocate a large enough buffer for the maximum string length permitted. When the data is greater than the buffer size, the Winlogon service terminates abnormally and your computer generates the error message. \n=====\nTry to use updated service packs, or check your system for Viruses. try to reinstall win xp with Repair Option,
186692
Use Write to save values and Input to read them.\n \n\nPrivate Sub cmdLoad_Click()\nDim fnum As Integer\nDim i As Integer\nDim value As String\n\n    ' Open the file.\n    On Error GoTo LoadError\n    fnum = FreeFile\n    Open "animals.dat" For Input As fnum\n    \n    ' Write the values.\n    For i = txtAnimal.lbound To txtAnimal.ubound\n        Input #fnum, value\n        txtAnimal(i).Text = value\n    Next i\n    \n    ' Close the file.\n    Close fnum\n    Exit Sub\n    \nLoadError:\n    MsgBox "Error" & Str$(Err.Number) & _\n        " loading data." & vbCrLf & _\n        Err.Description\nEnd Sub\n\nPrivate Sub cmdSave_Click()\nDim fnum As Integer\nDim i As Integer\n\n    ' Open the file.\n    On Error GoTo SaveError\n    fnum = FreeFile\n    Open "animals.dat" For Output As fnum\n    \n    ' Write the values.\n    For i = txtAnimal.lbound To txtAnimal.ubound\n        Write #fnum, txtAnimal(i).Text\n    Next i\n    \n    ' Close the file.\n    Close fnum\n    Exit Sub\n    \nSaveError:\n    MsgBox "Error" & Str$(Err.Number) & _\n        " saving data." & vbCrLf & _\n        Err.Description\nEnd Sub\n\nPrivate Sub cmdClear_Click()\nDim i As Integer\n\n    For i = txtAnimal.lbound To txtAnimal.ubound\n        txtAnimal(i).Text = ""\n    Next i\nEnd Sub\n\n \nNote that the Write method puts quotes around the values so the file looks like this:\n\n    "Ape"\n    "Bear"\n    "Cow"
750081
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !
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)
85088
U must have given the name of the variable in the "  ".\nGive it outside double quotes.
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