Anti-derivative of Log(x) is what?\n\nI know that the derivative of Log(x) is just 1/x.
214838
You have to use integration by parts.\nLet u = ln(x) and dv = dx, then\ndu = (1/x)dx and v = x\nint(udv) = uv - int(vdu)\n= xln(x) - int(x(1/x)dx)\n= xln(x) - int(dx)\n= xln(x) - x + C, where C is a constant
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
516867
1/5 log3 (x+3)\n\nExponentiate  (base 3)\nInverse Property\nDivide \n\n1/5  ( 3^log3(x+5))\n1/5  ( x+5)       (3^log base 3 and cancel each other out)\n\n(X+5) / 5\n\n(( I havent done this in a while, it may be wrong ))
526181
1.  f(x) = log[base3](3 - 2x), use the chain rule.\nf'(x) = (1/[(3-2x)ln3]) (-2)\n\n2.  y = ln |secx - tan x|\ny' = 1/(secx - tanx) (sec(x)tan(x) - (sec(x))^2\ny' = sec(x)\n\nI'd help you with the rest but I firmly believe if you truly needed help, you wouldn't have put all of those questions to begin with.  I highly advise you truly try learning instead of having people do homework for you.
658462
The problem is not that you can't take the log of a complex number. The problem is that log is multiple valued in the complex plane. In other words, the exponential function is not one-to-one. You can take the 'principle branch' of the logarithm, but then you can lose certain of the rules of logarithms.\n\nThe upshot is that when you take the log of -1, you either get an infinite set of values (.., -pi*i,  pi*i, ...) and another infinite set of possibilities for the log of 1 OR you lose the property that log(a^b)=b*log(a) that you used in your first paradox. This is just a fact of life when working with complex numbers. In the second, BOTH -pi*i and pi*i are logarithms of -1, so the argument fails. It's somewhat like saying that the squares of -1 and 1 are the same, so -1=1.
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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
59486
f(x)=Int(f'(y), 0, x)-f(0) => yes, they're opposite operation, but if x has a given value, e.g. x=2, 3...no, they're not, b/c in that case, the RHS of the previous egality is a real number, not a function
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For #1, there's no way you can solve that directly.  Note that it can be split up into three fractions, though.\n\n(x^2 + x + 1)/x = x^2/x + x/x + 1/x\n\nNow, we can actually reduce that, to\n\nx + 1 + 1/x\n\nWe solve for the antiderivative using the reverse power rule for derivatives.  That is, antiderivative for x^n is [x^(n+1)]/(n+1).  The exception is 1/x, where the antiderivative is ln|x|.\n\nSo the antiderivative of x+1+1/x is\n\n(x^2) / 2 + x + ln|x|\n\nBut we want the general antiderivative, so we add a constant C.\n\n(x^2)/2 + x + ln|x| + C
503501
(log x)^3 = log (x^9)\n(log x)^3 = 9 log x\n(log x)^2 = 9\nlog x = +/- 3\nx = 10^3 = 1000  or x = 10^-3 = .001
210062
pablodiablo is right, you can't treat an inequality like an equal sign when dividing or multiplying both sides by a negative number...swap the sign when doing so.  if you look at the graph of a logarithmic function, if 0<n<1, then log(n) is a negative number, and since y>x, x/y is less than one.  also, x & y must both be the same sign (positive or negative), otherwise x/y will be negative, and log of a negative number is undefined.
505878
e^(xlnx )=x^x\nderivative is  1+ln x
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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.
327345
First learn to put brackets properly.\nI presume that f(x) = [(x-1)*{(3x-2)^(1/5)}]/{(5x-3)^4}\nLet f(x) = y.\nTaking logarithm of both sides w.r.t. base 'e'\nln(y) = ln(x-1) + 0.2*ln(3x-2) - 4*ln(5x-3)\nTaking derivatives of both sides w.r.t. 'x'\n(1/y)*(dy/dx) = 1/(x-1) + 0.6/(3x-2) - 20/(5x-3)\nor dy/dx = {1/(x-1) + 0.6/(3x-2) - 20/(5x-3)}*f(x) [since y =f(x)]\nThus at x = 1, \ndy/dx = {1/(1-1) + 0.6/(3*1-2) - 20/(5*1-3)}*f(1)\n\nNow, 1/0 is undefined. So at x=1 the function f(x) does not have any defined derivative.\nHowever, if the term (x-1) doesn't occur in the expression for f(x) then \nf'(1) = (dy/dx)at x=1 = {0.6/(3*1-2) - 20/(5*1-4)}*f(1)\n= {0.6/(3*1-2) - 20/(5*1-3)}*(1/16)\n= - 47/80\n\nSo you must have copied the question wrongly.
518575
B is false: Let a be any number other than zero. If a>0, 2(2a)-a > 0, and f'(2a) = 2. Conversely, if a<0, 2(2a)-a<0, and f'(2a) = -2. Thus, any value of a other than zero serves as a counterexample.\n\nA is also false, although I'm not sure your teacher knows this. She may be thinking of something like f(x)=[a, b]∫g(t) dt, which is alawys a constant function regardless of g(t), and thus its derivative is zero. However, if the free variable appears in the limits of integration, the function may be nonconstant: for instance, let f(x) = [1, x]∫1/t dt, which is equal to ln x for positive values of x, and its derivative is certainly not zero. So the correct answer here is d: both A and B are false.
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1.52582219663167... if the log is base e\n1.59874065990736... if the log is base 10\n\nAs the use of a graphing calculator does not give any insight into the derivation of the solution, here it is:\n\nTo solve this, you need to use some numerical interpolation method, as the equation can not be solved directly for x (in terms of finitely many algebraic functions).\n\nNewton's method yields:\nLet y = f(x) = log(x) + 3x - 5 \n{here log(x) is the natural log(base e).  We want to find the value of x for which f(x) = 0}\nf '(x) = 1/x + 3\nthe method says that x(n+1) = x(n) + (y - f(x(n)) / f '(x(n))\nand here, y = 0, so\n\nx(n+1) = x(n) + (5 - log(x(n)) - 3x(n)) / (1/x(n) + 3)\n\nthen, x = lim(n→∞) x(n)\n\n(You must choose something reasonably close for x(1), such as\nx(1) = 1.  It takes only five iterations to get the accuracy of my solutions above in a VB6 program using double precision variables.  i.e., x(5)=1.52582219663167)
379299
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !
149553
Taking natural logarithm to both sides\nLn(y)= Ln(x) * Ln(x) = (Ln(x))^2\nDifferentiating\n(1/y )dy/dx = (2/x) * (Ln(x))\ndy/dx = (2y/x) * (Ln(x))\nat the point (e,e),\ndy/dx = (2e /e) * Ln(e)\nso the slope = 2