Find derivative of a natural log exponential function?
965203
y = (ln x²)^(2x+3) = e^((2x+3) ln ln x²)\nNow just ruthlessly apply the chain rule:\ndy/dx = e^((2x+3) ln ln x²) * (2 ln ln x² + (2x+3) * 1/ln x² * 1/x² * 2x)\nAnd simplify:\n(2 ln x)^(2x+3) * (2 ln (2 ln x) + (2x+3)/(x ln x))
173783
I believe that the integral of 1/x is equal to the natural log of x + C.  (ln x + C).  Usually you take the antiderivative of the function but there are exceptions to the rule, like 1/x.
658763
Take the logarithm of both sides:\n\nlog a^a = log 2\n\nThen use the rule log (x^y) = y log x\n\nso a log a = log 2\n\nUnfortunately from here we can only solve numerically. Look at the graph of a log a and make approximations.\n\nThough not expressible using elementary functions, the actual answer is defined by the Lambert's W function and reads: log 2 / (LambertW (log 2)), which is roughly 1.5596. More about the W function and what it's used for below.
347784
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
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
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)  )
1035877
The natural logarithm (over the real numbers) is defined only for positive numbers, so \n\nx-4>0\n\nin other words\n\nx>4.
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
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)
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
179378
All you do is take the natural logarithm (ln) of both sides.\nBecause the ln of e to any power is just that power it becomes this:\nln (e^-x) = ln(6) \nThe ln sort of "cancels out" the e, leaving you with the power so...\n-x = ln(6)\nMultiple both sides by negative 1 and you're done, unless you need to approximate ln(6).
1130920
You can't solve this using traditional techniques since you'll either have a logarithm or an exponential on one of the x variables.  The people claiming Taylor series and techniques like Newton-Raphson approximation are most likely on the right track.  I would use a CAD program, or graph the function y(x)=(1+x)^2-2^x and find the zeroes that way by zooming and such (on at TI-83 or similar).\n\nIn this case, I'll use MathCAD.  The solutions are:\nx={-1.57862, 0, 5.31972} (two of these are approximate, by the way)
826204
Do you mean \n\nlog (base n) (x) = 1/(log (base x) (n))\n\nWell let log (base n) (x) = a\nTherefore x = n^a\n\nTake logs (base x)\nSo 1 = log (base x) (n^a)\n= a * log (base x) (n)\nBut a = log (base n) (x)\nTherefore log (base n) (x) * log (base x) (n) = 1\nie log (base n) (x) = 1/(log (base x) (n)) ............. QED
1052815
3.1 - Given the equation of an exponential function, state its domain and graph it. \n\n3.2 - (a) Given a logarithmic equation, express it as an exponential equation.\n(b) Given an exponential equation, express it as a logarithmic equation. \n\n3.3 - Given the equation of a logarithmic function, state its domain and graph it. \n\n3.4 - (a) State the product, quotient and power properties for logarithmic functions.\n(b) Apply the properties of logarithms to expand and/or simplify expressions involving logarithms. \n\nSolving Exponential and Logarithmic Equations\n- Using the laws of exponents and the properties of logarithms, simplify expressions involving the composition of exponential and logarithmic function of the same base. \n - Given an equation involving logarithmic functions, solve it. \n\n - Given an equation involving exponential functions, solve it. \n\n- Given an applied problem described by an exponential or logarithmic equation, solve the
830737
I do not understand what it is you are trying to solve. Are yoiu trying to find the inverse fuction? If you are that will be extremely difficult. If you seek the derivative then it has been doone above for you.\n\nIf you seek the integral then it would probably requiire a power series But as it is your problem is not a problem as you have not stated what you want solved.\n\ny = x^x\n= e^xlnx\n\ndy/dx = (lnx + 1)x*x\n= (lnx +1)*y\nSo lnx = 1/y *dy/dx - 1\nie x = 1/e * e^(1/lny * dy/dx)\nWhence x = 1/e * e^ [1/(lny* dx/dy)]\n\nNow I find that unconvincing as to get dx/dy I need to know x in terms of y Its looped onto itself!!!! In fact it is a differential equation not a function. ( x = 1/e * e^ [1/(x' . lny)]
212357
As I am sure you soon realized, this is a bit challenging. You cannot solve it easily by algebra, although you can make a series expansion approximation of ln y, or exp(y) and approximate the answer to this kind of equation in some cases. \n\nA more general approach is apply some numerical technique, like the Newton-Raphson method. \n\nLet F(Y)=ln(Y) + 10Y -4\n\nYou want to find the roots of F(Y), i.e., those values Y* such that F(Y*)=0. \n\nSet up an iteration Yn+1 = Yn - F(Yn) / F'(Yn)\n\n\n\nwhere the derivative F'(Y)= (1/Y)+10. \n\nTry it and see if it converges for you.
746085
taking log on both sides\nlny=(x-5)ln2\n1/y(dy/dx)=ln2\ndy/dx=yln2=2^(x-5)ln2