[MOVED] Find the exponential function passing through points

[Guest]

Find the exponential function y=Ce^kt that passes through the points (3, 1/2) and (4, 5)

My work thus far....

1/2C = Ce^3k

1/2 = e^3k

ln(1/2) = ln(e^3k)

ln(1/2) = 3k

k = ln(1/2)/3

k = -0.231

If I use the other point (4, 5)

my k = 0.402

I thought k was constant. How do you go about getting the equation for these 2 points if k is different at each of the two points? Thanks

 

[tkhunny]

Re: Find the exponential function

1/2C=Ce^3k

Whence came that 'C' on the left hand side?

1/2 = Ce^(3k)

5 = Ce^(4k)

Now try it.

 

[Guest]

Now I am unsure how to do this at all.

what am I solving for in 1/2=Ce^3k

k? or C?

 

[tkhunny]

You have two equations and two variables. Your task is to solve them simultaneously.

1/2 = Ce^(3k) ==> C = (1/2)*e^(-3k) <== An expression for 'C' in terms of 'k'.

5 = Ce^(4k) <== Substitute that in here.

5 = ((1/3)*e^(-3k))e^(4k)

You should be able to solve that for 'k'. Once you have 'k', you can go back and find 'C'.

Let's see what you get.

 

[Guest]

So, here is what I get.

y=.0005e^2.303t as my exponential function.

Please tell me if that is correct. And, I solve for C and substitute that into my other equation so everything is in terms of k?

Thanks again!!!

 

[soroban]

Re: Find the exponential function

Hello, ezrajoelmicah!

Find the exponential function \(\displaystyle y\:=\:Ce^{kt}\)
that passes through the points \(\displaystyle \left(3,\,\frac{1}{2}\right)\) and \(\displaystyle (4,5)\)

Plug in the given values and solve the system of equations . . .

At \(\displaystyle \left(3,\,\frac{1}{2}\right)\), we have: \(\displaystyle \:\frac{1}{2}\:=\:Ce^{3t}\;\) [1]

At \(\displaystyle (4,5)\), we have: \(\displaystyle \:5\:=\:Ce^{4t}\;\) [2]

Divide [2] by [1]: \(\displaystyle \L\:\frac{Ce^{4t}}{Ce^{3t}}\)\(\displaystyle \:=\:\frac{5}{\frac{1}{2}} \;\;\Rightarrow\;\;e^t\:=\:10\;\;\Rightarrow\;\;\L t\:=\:\ln10\)

Substitute into [2]: \(\displaystyle \:Ce^{^{4\cdot\ln10}}\:=\:5\;\;\Rightarrow\;\;Ce^{^{\ln10^{^4}}}\:=\:5\;\;\Rightarrow\;\;C\cdot10^4\:=\:5\)

. . Hence: \(\displaystyle \:C\:=\:\frac{5}{10^4}\;\;\Rightarrow\;\;\L C\:=\:0.0005\)


The function is: \(\displaystyle \:y\;=\;0.0005e^{^{t\cdot\ln10}} \;=\;0.0005e^{^{\ln10^{^t}}}\)

. . Therefore: \(\displaystyle \L\:y\;=\;0.0005\left(10^t\right)\)

 

[tkhunny]

I have met a few who hate it when you get rid of the 'e'. It makes sense to me, when it is logically motivated, but some folks just love to see that 'e' in the final result. It's just a style difference. If your teacher insists on an 'e', just do it that way and smile inside, knowing it's just not necessary.