It is a number such that, if 10 were raised to that power, the answer would be k.
ie 10log k = k
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∙ 13y agoy = c ekxpassing through (1, 180) and (3, 20).(Note: All of the 'log' in the following are natural logs.)log(y) = log(c) + k log(x)Log(180) = log(c) + k log(1) = log(c)c = 180log(20) = log(180) + k log(3)log(20) - log(180) = k log(3)k = log(20/180) / log(3) = - log(9) / log(3) = -2y = 180 e-2xIn checking my work, I find that this doesn't work at all.Oh woe! Where have I failed ?--------------------------------------------Here's what I did:y=Cekx which passes through the points (1,180) & (3,20)(1) Substitute the values of x & y to create two equations:180 = Cek & 20 = Ce3k(2) Rearrange in terms of the constant, C:C=180/ek & C=20/e3k(3) Since both sets of points satisfy the exponential function, the constant will be the same so:180/ek = 20/e3k(4) Now rearrange and solve for k:180e3k = 20ek9 = e(k-3k)ln(9) = ln(e-2k)ln(9) = -2kk = -ln(9)/2 which is approx. -1.09861(5) Now substitute k into one of the equations to solve for C:C =180/e(-1.09861)C = 539.9987 or rounded to 540So the exponential function that includeds the points (1,180) & (3,20) can be approximated as:y = 540ekx , where k = -1.09861
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The graph of log base b(x-h)+k has the following characteristics. the line x = h is a vertical asymptote; the domain is x>h, and the range is all real numbers; if b>1, the graph moves up to the right. of 0>b>1, the the graph moves down to the right.
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
how do i log in
k=log4 91.8 4^k=91.8 -- b/c of log rules-- log 4^k=log 91.8 -- b/c of log rules-- k*log 4=log91.8 --> divide by log 4 k=log 91.8/log 4 k= 3.260
If in the real number universe, first k is to be >0, y=kx = exlog(k) the antiderivative of eax is eax/a so the antiderivative of Y is exlog(k) / log(k) = kx /log(k)
You could just use the binomial theorem. Step through rows, n, and entries, k, and compute the Pascal's triangle value as n!/(k!*(n-k)!) You'll actually have better luck if you use the natural log of a factorial, then you can use laws of exponents to get: exp(log(n!/k!/(n-k)!)) = exp(log(n!)-log(k!)-log((n-k)!)) = exp(logfact(n)-logfact(k)-logfact(n-k)) which won't run into the integer overflow problems that a plain factorial function would have. To fill up a logfact array, something like this might work: while(i<maxn) logfact(i)=logfact(i-1)+log(i) i=i+1 Wend Be careful to initialize correctly, and watch your conversion between integers and doubles (probably have to do some rounding to your final answers).
y = c ekxpassing through (1, 180) and (3, 20).(Note: All of the 'log' in the following are natural logs.)log(y) = log(c) + k log(x)Log(180) = log(c) + k log(1) = log(c)c = 180log(20) = log(180) + k log(3)log(20) - log(180) = k log(3)k = log(20/180) / log(3) = - log(9) / log(3) = -2y = 180 e-2xIn checking my work, I find that this doesn't work at all.Oh woe! Where have I failed ?--------------------------------------------Here's what I did:y=Cekx which passes through the points (1,180) & (3,20)(1) Substitute the values of x & y to create two equations:180 = Cek & 20 = Ce3k(2) Rearrange in terms of the constant, C:C=180/ek & C=20/e3k(3) Since both sets of points satisfy the exponential function, the constant will be the same so:180/ek = 20/e3k(4) Now rearrange and solve for k:180e3k = 20ek9 = e(k-3k)ln(9) = ln(e-2k)ln(9) = -2kk = -ln(9)/2 which is approx. -1.09861(5) Now substitute k into one of the equations to solve for C:C =180/e(-1.09861)C = 539.9987 or rounded to 540So the exponential function that includeds the points (1,180) & (3,20) can be approximated as:y = 540ekx , where k = -1.09861
emagrants
it is used as a way of measuring how fast cells are dividing, defined as the doubling rate, and it is worked out with the following formula: k=(Log Nt - Log No)/ t x Log 2 this goes to the slightly easier form of; k= 3.32 x (Log Nt - Log No)/ t where k= growth rate constant Nt = number of bacteria at second time No = number bacteria at start t = time gone. (obvioulsy you take the logs of Nt and No in the formula)
p;kp;k;ko;jk;
The graph of log base b(x-h)+k has the following characteristics. the line x = h is a vertical asymptote; the domain is x>h, and the range is all real numbers; if b>1, the graph moves up to the right. of 0>b>1, the the graph moves down to the right.
gt is the pagal log pagal ho tm log me khe raho hu k zoroastrianims ki holy book kya hai
It is measured by means of wave motion and by a logarithmic function of the form R = k log I.
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If delta G is negative, then K (Upper case K, as in Keq or the equilibrium constant) will be greater than 1. Remember that delta G = -RT log K.Do not get Keq confused with lower case k, which denotes rate constants (which have NOTHING TO DO WITH Keq or delta G).