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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
What does Ln mean?
in math, ln means natural log, or loge and e means 2.718281828
Where can i find my math problems from my math book online?
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.
How do you solve log x3?
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.
Hod do you register for math blaster?
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.!
How do you solve for a power in a math problem?
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)
How do you log into st math with a log in?
What is it
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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log 8.008
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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.
How do you do log math problems?
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
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Exponent, expression, e (natural log)
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I don't know but mine says the same thing!
Why cant you log your school in to my maths?
1) Problem with your schools log-in. 2) talk to the person that created the my maths account.
What does Ln mean?
in math, ln means natural log, or loge and e means 2.718281828
