arrange in ascending order of growth rate  f1(n) = 10^n  f2(n) =n^ ⅓  f3(n) =n f4(n) = log2n f5(n) = 2^ log n
830317
f1>f3>f2>f5>f4\nJust replace the variable n in  Excel's function and compare the result.
694817
f-8-3f = 3-7f+4\nf-3f+7f = 8+3+4\n8f-3f = 15\n5f = 15\nf=15/5\nf=3
883446
Hi, it depends on particular organisms. Check the link below for basic mathematical formulas you may be able to use.\n\nA study is done in which the number of bacteria in a liquid culture is monitored over a period of several hours. Ten observations were made:\n\n\nTwo questions need to be answered:\n\nIs this data consistent with an exponential growth model: N = N0ekt, where \nN0 = the number of bacteria present initially (at t = 0)\nk = a growth rate constant\nN = the number of bacteria present at a given time t,\nand e is the fundamental mathematical constant forming the base for natural logarithms. \nIf the answer to the first question is "yes", then what are the values of N0, k, and t2 (the so-called doubling time) for this bacterial system?\nThe experimental measurements listed above indicate the number of bacteria are growing with time. If we plot the observed values of N (vertically) against the elapsed time, t, in hours, we get a graph (available on web page below).\n\nYou can draw a smooth curve through the experimental points as shown in this figure, and the points all seem to lie relatively close to this curve. Further, this curve looks quite a bit like the graph of an exponential function, but from what we see, it might just as well be part of a parabola or of one of a number of other curves that result when various algebraic forms are graphed. This graph suggests that the answer to the first question may be "yes", but we would usually want stronger evidence than this to draw a definite conclusion. The trouble here is that it is difficult to distinguish precisely and reliably between the curves are the graphs of quite different mathematical functions.\n\nThere is, of course, one curve that is easily recognized by everyone - the straight line! In this situation, the logarithmic function can be used to rewrite the exponential growth formula into a form that will allow us to create a plot that does give a straight line when data is the result of an exponential growth (or decay) process. Start with the exponential growth formula:\n\nN = N0 ekt\n\nSince this involves an exponential with base e, it is "natural" to use natural logarithms - those based on the number e. Using the more common notation ‘ln’ for ‘loge’, we simply take the logarithm of both sides of the previous equation:\n\nln N = ln(N0 ekt)\n\n= ln N0 + ln(ekt)\n\n= ln N0 + kt ln e\n\n= ln N0 + kt\n\nHere, we have used two of the three basic properties of logarithms to manipulate the right-hand side, namely that\n\nlog (AB) = log A + log B \n\nlog (Am) = m log A \n\nFurther, in the last step we took advantage of the fact that by choosing to work with natural logarithms, ln e = logee = 1. Thus, the exponential growth formula can be written in the logarithmic form\n\nln N = kt + ln N0\n\nIf the exponential growth formula is correct and we were to plot ln N along the vertical axis of our graph (rather than N itself), and t along the horizontal axis, then the equation of the plot we are preparing will have the generic form\n\nvertical quantity = constant x horizontal quantity + another constant \n\nBut this is just the equation of a straight line. The plot that results using the data given above is available on web site below.\n\nAlthough the values written on the vertical axis are values of N, this graph is ruled vertically so that the vertical distances are actually proportional to ln N. Since one scale is logarithmic (the vertical one) whereas the other graph scale, the horizontal one, is the ordinary uniform spacing type, this is called a ‘semilogarithmic plot’.\n\nAt any rate, the figure indicates that the points are scattered about a straight line, the scatter due to small random errors in measurement. Since the plot of ln N vs t gives a straight line, we have confirmed that\n\nln N = kt + ln N0 (equation *) \n\nis a valid formula for this data, and therefore, so is its equivalent, N = N0 ekt, valid for this data.
754998
a 4\nb 5\nc 3\nd 2\ne 3\nf 4\ng 3\nh 5\ni 5\nj 5\nk 3\nl 5\nm 5\nn 5
893153
2Fe (s) + 3F2 (g) ---> 2FeF3(s)
899734
β=10 log I / Iο\nβ2=10 log I2 / Iо\nβ2-β=10 log I2 / Iо - 10 log I / Iо\n         =10 log I2 / Iо / I / Iо\n         = 10 log I2 / I\n         = 10 log 2 I / I\n         = 10 log 2 = 10 x o.3 = 3\n\nI hope i helped you, sorry if i have done any mistake.
704007
use arithmatic progression formula n(2a+(n-1)d)/2\nwhere n=no. of terms. a=value of first term and d=common difference between each consecutive term\nsay fro your first eg. no. of black ball = 1+3+5+...\nn=10 a=1 d=2\nso sum = 10(2+9*2)/2 = 100\nsolve the rest by changing values of n, a and other variables respectively
406750
6*5*4*3 = 360 if order matters.\n\n6*5*4*3 / 4! = 15 if order does not matter
1031756
It would help if you used some parentheses to indicate which terms are being divided into which terms!  I suspect that just using the order of precedence for operations will not result in the problem you are actually trying to solve.
346393
all are the same\n\n10*10^5=\n1*10^6=\n2*10^5.7=\n4*10^5.4=\n100*10^4\n\ndecibels use a logarithmic scale so you have to compare them as exponents
114069
log(a*b) = log(a)+log(b)\n\nlog(a/b) = log(a)-log(b)\n\nSo these work when expanding (left to right) and condensing (right to left).\n\n\nThanks for the update: So using your example:\n\n2ln3 - ln5 = ln(3^2) - ln5 = ln9 - ln 5 = ln(9/5)\n\nThe reason i converted 2ln3 to ln9 is because the condensing form only works when the coefficients of the logs are identical (ie 1 in this case). The formula for doing this is N*ln(x) = ln(x^N)\n\nNext, \n\nln(7X/3) = ln7 + lnX - ln3 (or ln(7/3) + lnX)\n\nAs you can see, the above form at 3 factors inside it - the expansion form will work for however many factors you want to use.\n\nIf you plug in the numbers into a calculator then you will see that this works.
367516
here are the anwsers in order from top to bottom \n-T\n-T\n-F\n-F \nhope this helps
559365
f(x)=x+3\nif x=0, f(0)=0+3=3\nif x=2, f(2)=2+3=5\nif x=-2, f(-2)=-2+3=1\n\nfor g(x)=2x+1\nif x=0, g(0)=2*0+1=1\nif x=2, g(2)=2*2+1=5\nif x=5, g(5)=2*5+1=11\n\nfor h(x.y)=2x+y\nif (x,y)=(0,0) h(0,0)=2*0+0=0\nif (x,y)=(2,1) h(2,1)=2*2+1=5\nif (x,y)=(1,3) h(1,3)=2*1+3=5
96099
Your anwser is incorrect.  f (a+1)=3(a+1)+3 First distribute; 3a+3+3 so the anwser is 3a+6!
743256
First 4*log[x-2] = log[x-2]^4. \nThen -log[5] = log[1/5]\n\nThe sum of logs is the log of the product of the arguments and so this gives\n\n(1/2)*log[(x-2)^4 * (x+3) / 5]\n\nFinally the (1/2) becomes an exponent in the argument:\n\nlog[((x-2)^4 * (x+3) / 5)^(1/2)]
679340
By definition, you would do the following:\n\nf'(x) = lim(h->0) (f(x+h) - f(x))/h\nf'(x) = lim(h->0) ((3(x+h) - 3x)/h\nf'(x) = lim(h->0) 3h/h\nf'(x) = lim(h->0) 3\nf'(x) = 3\n\nOr, you could simply do use the following fomulas:\n\nd/dx[af(x)] = a(d/dx[f(x)])\nd/dx[x^n] = nx^(n-1)\n\nd/dx[3x] = 3(d/dx[x]) = 3(d/dx[x^1]) = 3*(1*x^0) = 3\nf'(x) = 3
805545
Part A\nlog(base3) 20 = (log 20) / (log 3)\nlog 20 = log (2*10) = log 2 + log 10 = 1.301\nlog 3 = 0.477\n1.301/4.77 = 2.727 --> about 2.73\n\nPart B\nlog(base 9) 100 = (ln 100) / (ln 9)\nln 100 = ln (10^2) = 2 (ln 10)\nln 9 = ln (3^2) = 2 (ln 3)\n(ln 10) / (ln 3) = 2.30 / 1.10 = about 2.10\n\nPart C\n1) log(base a) 6xy^5z^4 = log(base a) 6 + log(base a) x + log(base a) y^5 + log(base a)z^4 = log(base a) 6 + log(base a) x + 5 log(base a) y + 4 log(base a) z\n\n2) log (5a/4b^2) = log 5a - (log 4b^2) = log 5a - (log 4 + log b^2) = log 5a - log 4 - 2 log b\n\nPart D\n(1) 1/2 log a = log (a^(1/2))\nlog (a^(1/2)) - log 2 = log (a^(1/2) / 2)\n\n2) ln 2x + 3 ln x - 3 ln y = ln 2x + ln x^3 - ln y^3 = ln (2x*x^3)/y^3\n= ln (2x^4/y^3)\n\npardon me if you see typos.\nIt's 1:30 in the morning