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.
YOU GO MATH BLASTER YOUR LOGIN NAME WILL BE ON THE LEFT then at the right top corner click log out next at the same corner click register and the sreen will have the register stuff for you and that is how you you register for math blaster. Your Welcomw Everybody who needs help registering for math blaster.!
You have to use logarithms (logs).Here are a few handy tools:If [ C = D ], then [ log(C) = log(D) ]log(AB) = log(A) + log(B)log(A/B) = log(A) - log(B)log(Np) = p times log(N)
What is it
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
log 8.008
between 180-200
A fault log is used to record all details of a problem as well as the client details and the resolution to the fault/problem.
Sometimes you need to take logs, or antilogs, on both sides of an equation. Sometimes you need to apply certain common logarithmic identities, especially: log(xy) = log x + log y log (x/y) = log x - log y log (ab) = b log a
Exponent, expression, e (natural log)
the system log record (page 698)
in math, ln means natural log, or loge and e means 2.718281828
I don't know but mine says the same thing!
1) Problem with your schools log-in. 2) talk to the person that created the my maths account.