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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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
in math, ln means natural log, or loge and e means 2.718281828
For my school, you can go on a web site called Classzone.com and when you first get on you can pick a subject and state. The site will search for your book but you have to have a school code to first log on. If your school does have this site then you log on with the school username and password then make one of your own as you go on.
You can't solve this since it isn't an equation.There is also an ambiguity (it's hard to write math on a typewriter keyboard) - are we talking about log(x3) or maybe logx(3)?Restate the question: Simplify log(x3)Answer: 3log(x)You could explain this by saying: log(x3) = log[(x)(x)(x)] = logx + logx + logx = 3logx. The general rule is log(xn) = nlogx.
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