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180 10 log(x) 130 = -(117000 i integral_(-iinfinity+gamma)^(iinfinity+gamma)(Gamma(-s)^2 Gamma(1+s))/((-1+x)^s Gamma(1-s)) ds)/pi for (-1<gamma<0 and |arg(-1+x)|<pi)

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Q: What is the answer to this math problem 180 10 log x 130?
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What is the exponential function in the form y equals c times e to the power of kx that passes through the points 1 180 and 3 20?

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


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What is the exponential function in the form y equals c times e to the power of kx that passes through the points 1 180 and 3 20?

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


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