What is the anti-derivative of Log(x) or even Ln(x)?
473554
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
773933
1-In general the derivative of ln|U| is U'/U. In your case U = 2x^x + 4x + 1. Using the fact that x^x = exp(xlnx) you can easily check that the derivative of \n\nU' = 2(1 + lnx)x^x + 4.\n\nThus the derivative of your function is\n\n[2(1+lnx)x^x+4]/[2x^x+4x+1].\n\n2- The derivative of U^a, where a is a real number, is aU^{a-1}U'. In your case U = x e^x and a=1/2. Since U' =(x+1)e^x, one has\n\n(sqrt(xe^x))' =1/2{(x+1)/sqrt(xe^x)}e^x = 1/2{(x+1)/sqrt(x)}exp(x/2).
898977
log(2x+1)-(1/3)log(x+1)\n=log(2x+1) - log(  (x+1)^(1/3)  )\n=log( (2x+1) /   (x+1)^(1/3)  )
948694
The anti-derivative of a function f(x) is another function F(x) so that F´(x) = f(x). \n\nIf f(x) = x^n (n<-1 or n>-1) then the anti-derivative is F(x) = (x^(n+1))/(n+1)  + C ... we need to add an arbitrary constant.\n\nIn your case we have:  f(x) = x^1  and  F(x) = (x^2)/2 + C
221721
Product rule:\n\n(fg)' = (f ')g + f(g ')\n\nSo,\n\n(-2x^2 * ln(x))' = (-4x * ln(x)) + (-2x^2 * 1/x)\n\nMultiply and clean up:\n\n-4xln(x) - 2x
470727
log_7(7)+log_7(x)=1+log_7(x).
962408
log4+xlogb=xloga\nx(loga-logb)=log4\nx=log4/(loga-logb)
1085682
Ln(e^x+1) = 0 by log laws this equals X+1 Ln e = 0 and Ln e cancels therefore X+1 = 0, x = -1\n\nIf the equation does not = 0 then you will be just left with ( x + 1)
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.
533385
Please believe TheoDorkx. He's the only one who understands logarithms.\n\nAnother way to solve logarithm problems is directly change from the logarithmic to the exponential form.\n\nln(2x) = 2 means that 2 is the power to which you raise the base of the logarithm, e, in order to get the result 2x.\n\nSo the equivalent exponential form is\n\n2x=e^2\n\ntherefore\n\nx=(1/2)e^2\n\nThis is the same thing TheoDorkx got with his approach. I offer it as a slightly different way to think about the problem.
1086154
The reason we call ln(x) "natural" is twofold: first, expressions in which the unknown variable appears as the exponent of e occur much more often than exponents of 10 (because of the "natural" properties of the exponential function which allow it to describe growth and decay behaviors), and second, because the natural logarithm can be defined quite easily using a simple integral or Taylor series--which is not true of other logarithms.
488973
If you mean this (e^x)^x, then that is e^(x^2).  There is no closed form expression for the antiderivative of this with respect to x.\n\nHowever if you mean e^(x^x), then I don't know.\n\nUPDATE: Hey discoganya, I'm getting d/dx(x^x) to be \n(x^x)(ln x + 1), not x^x.  Now (d/dx)e^x = e^x.\n\nUPDATE (again): I think that part of the problem with that, discoganya, is that you are assuming that x is a whole number.\n\nThis is how I got my dervative.\n\n(d/dx)(x^x) = (d/dx)(exp(xln x))\n= exp(xln x)(1*ln x + x(1/x))\n= (x^x)(ln x + 1), for x > 0\n\nI have run this through Mathematica, and it confirms it.\n\nI believe the only function such that f'(x) = f(x) is f(x)=  a*exp(x), where a is an arbitrary constant.\n\nBy the way, Mathematica also could not find the integral of e^{x^x), honey, so it also probably does not exist in a closed form.
852324
We have f(x) = (x)^(7x) = e^(7x*ln(x)). So, f'(x) = e^(7x*ln(x)) * d/dx(7x*ln(x) = . (x)^(7x) *7 *(x *1/x + ln(x) = 7* (1+lnx) * (x)^(7x)
1075997
Since you use log rather than ln, I assuming base 10 (common) logs.\n\nLogrithms are exponents!!!!!\nThey are the exponent you hang on the base to get the number in question.  That is,  saying   log x = y is the same as saying 10^y = x.\n\nSo log x = -2 is the same as saying 10^(-2) = x  or x = 1/100.
519118
you use the 10^x key\n\nthe antilog of -4.82 \n= 10^(-4.82) = 1.514 * 10^(-5) = 0.00001514\n\ntype in -4.82 and hit the 10^x key (it's usually right above the log key, which mean you hit the 2nd function key then log)
691996
Let u = lnx, then du = (1/x)dx\nln(e) = 1 and ln(infinity) = infinity\nint from 1 to infinity of 1/(u*sqrt(u)) du\nu*sqrt(u) = u*u^(1/2) = u^(3/2)\nint from 1 to infinity of 1/u^(3/2) du\n= int from 1 to infinity of u^(-3/2) du\n= limit as b goes to infinity of -2u^(-1/2) from 1 to b\n= limit as b goes to infinity of -2(b^(-1/2)-1)\n= -2(0-1)\n= 2
956834
For log(x+8) to be log x +log 8 ,x+8 must be equal to 8x\nWe can write it algebrically as log(x+8)=logx+log8=log(8x)\n=>8x=x+8\n=>7x=8    or,x=8/7