How to solve Exponential eqations?
830296
The equation is :\nI = Ia 10^-kx\n\na) if x =50 then I = 0.2 Ia \nSo 0.2 = 10^-k50\ntaking log decimal :\nlog(0.2) = -k50    so k = -log0.2 /50 = (log5) /50 = 0.0140 m^-1\n\nb) if x= 20m\nI = Ia 10^-kx = Ia 10^-0.0140*20 = 0.52 Ia\nSo the percentage is 52%\n\nc) If the intensity is the half I = 0.5 Ia\n10^-kx = 0.5\nSo x = log 2 /k = 21.5 m
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
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.
217977
f(E) = 1/(exp{(E-Ef)/kT}+1)
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
970181
1) This has no elementary solution\n2) This one, however, does have a solution, with the substitution e^x = tanu.  You are at the answer arctan(e^x)\n\nSteve
895385
"Give a student an answer, and their grade on that assignment will go up a point, teach them how to find their answer and their score on the final exam will improve!"\n\nRadioactive decay is an exponential thing; for every 24000 years that elapses in this problem, half the remaining material vanishes.  If you express this as an algebraic equation you can figure this stuff out easily.  The hard part is figuring out the equation.\n\nLets say you start out with X_0 grams of material.  The amount of material remaining is X: \n\nX= X_0*(1/2)^(t/24000)\n\nWhere t is given in years.  See how this works?  if t = 0, then X=X_0*1/2^0 power.  Anything raised to the 0 is always 1, so X=X_0.  At t=24000 years, X=X_0*1/2^(1), so X=X_0*1/2; half as much is left.\n\nIn this problem you are given t, and X and are asked to find X_0 so  you use a calculator to turn 1/2^(21500/24000) into just a number and then you have\n\n4 grams=X_0*number\n\nPretty easy to solve that for X_0.  The point is, once you write down the equation that describes the system, you can manipulate it to solve for any of the terms in it, as long as you know the other two terms.\n\nIncidentally, I get a different answer than those other two folks who agree with one another.  They are wrong, but not by a lot.  Work it out for yourself.\n\n[Edit:  In fact I get the same result IMI does, so my method does work. Her answer was not posted when I started working on mine.]
736107
If u = e^x, then 1/u = 1/e^x = e^(-x)\n\n(u - 1/u) / (u + 1/u) = -t\n\nMultiply both sides by (u + 1/u)\n\nu - 1/u = -t (u + 1/u) = -ut - t/u\n\nMultiply both sides by u\n\nu^2 - 1 = -u^2t - t\nu^2 + u^2 t = u^2(1 + t) = 1- t\nu^2 = (1 - t)/(1 + t)\nu = +/-sqrt[(1-t)/(1+t)]\n\nRe-substitute u=e^x:\n\ne^x = +/-sqrt[(1-t)/(1+t)]\n\nTake ln of both sides:\n\nx = ln[sqrt[(1-t)/(1+t)]\n\nSorry I screwed that up the first time.  I need sleep.  Thanks for getting it right, Sidarth.
793577
|e^x - 5| = e^x - 5 for e^x - 5 ≥ 0 ie x ≥ ln5\n= 5 - e^x for e^x - 5 < 0 ie x< ln5\n\nSo ∫|e^x - 5|dx from [-1,4]\n\n= ∫(5 - e^x) dx from [-1, ln5] + ∫(e^x - 5) dx ]\n\n= [5x - e^x] from [-1, ln5] + [e^x - 5x] from [ln 5,4\n\n= [(5ln5 - 5) - (-5 - e^(-1))] + [(e^4 - 20) - (5 - 5ln5)]\n\n= e^4 + e^(-1) - 15\n\n(≈ 34.97)
139157
Try to fit your data with Excel’s exponential fitting curves,\nIt will give you Coefficients which make the exponential looks like a line.\n\nOr try the experiment more rapidly.
492313
x^2*y"-x*y' +4*x*e^(x^3)=0\n\nx[x*y"-y' +4*e^(x^3)]=0\n\n=> x=0 or x*y"-y' +4*e^(x^3)=0\n\nx*y"-y' +4*e^(x^3)=0\n\n=>\n\n∫[x*y"-y' +4*e^(x^3)].dx=0\n\n=> [-4*x*Gamma[1/3, -(x^3)]]/[3(-(x^3))^(1/3)]=0\n\nwell even my computer couldn't solve that, so let the answer be x=0.
648708
I know this! I did a problem like this a while ago! It's an exponential equation. It can be used to predict exponential increases or decreases, like bacteria or double/triple/quadrupled information.
621223
Try the exp() function.  For more info visit http://users.ece.gatech.edu/~bonnie/book/TUTORIAL/tut_1.html
772997
A/B=e^(-t/5)\nln(A/B)=-t/5\n-t=5ln(A/B)\nt=-5ln(A/B)
728396
1.\nln(5x + 2) + ln x = ln e\nln(5x + 2) + ln x = 1\ne ^ (ln(5x + 2) + ln x) = e ^ (1)\n(e ^ ln(5x + 2))(e ^ ln x) = e\n(5x + 2)(x) = e\n5x^2 + 2x - e = 0\nx = (-2 +/- sqrt(4 - (4)(5)(-e))) / 2(5)\nx = (-2 +/- sqrt(4 + 20e)) / 10\nx = (-2 +/- (2)sqrt(1 + 5e)) / 10\nx = (-1 +/- sqrt(1 + 5e)) / 5\ntest answers to ensure ln of positive number\nx = (-1 + sqrt(1 + 5e)) / 5\n\n2. \nln(x + 1) / ln(x - 1) = 2\nln(x + 1) = (2)ln(x - 1)\nln(x + 1) = ln((x - 1)^2)\nx + 1 = (x - 1) ^ 2\nx + 1 = x^2 - 2x + 1\n0 = x^2 - 3x\nx(x - 3) = 0\nx = 0 or x = 3\ntest answers to ensure ln of positive number\nx = 3
815275
Using the quotient rule,\n\nlog(2x-1) - log(x-3)=1\nlog((2x-1)/(x-3)) = 1\n\nThen change to exponential form:\n\n10^1 = (2x-1)/(x-3)\n\nthen solve:\n\n10(x-3)=2x-1\n10x-30=2x-1\n8x=29\nx=29/8
352749
By the chain rule:\n[e^(3x^(2)-7x)]*(6x-7)