Log word problem: C=e(natural log ie 2.718)^-k(t+b)

[Posher]

So I have this question C=e(natural log ie 2.718)^-k(t+b) The initial concentration is 500 (where t=0) When t = 5 the concentration is 175

This is the shortened version if you want to see the rest I'll type it all out but you should get the gist

What is getting me stuck is I have to find k and b!!!

 

[JeffM]

So I have this question C=e(natural log ie 2.718)^-k(t+b) The initial concentration is 500 (where t=0) When t = 5 the concentration is 175

This is the shortened version if you want to see the rest I'll type it all out but you should get the gist

What is getting me stuck is I have to find k and b!!!

For goodness sake, please make the effort to transcribe your problem fully and completely.

If your function is \(\displaystyle C = e^{-k(t+ b)}\), your function does not even contain a natural log.

If that function above is correct, you can USE the natural log function to find b and k.

\(\displaystyle C = e^{-k(t+b)} \implies ln(C) = ln \left ( e^{-k(t+b)} \right ) = -\ k(t + b) * ln(e) = -\ k(t + b).\)

Follow that?

So now \(\displaystyle ln(500) = -\ k(0 + b)\ and\ ln(175) = -\ k(5 + b)\).

Now what?

 

[Ishuda]

So I have this question C=e(natural log ie 2.718)^-k(t+b) The initial concentration is 500 (where t=0) When t = 5 the concentration is 175

This is the shortened version if you want to see the rest I'll type it all out but you should get the gist

What is getting me stuck is I have to find k and b!!!

There are several ways to go about it. Possibly the easiest is
C = e^-k(t+b) = e^-kb e^-kt
or, since
500 = e^-kb,
C = 500 e^{-kt
Now use the value at t=5 to compute k, then compute b.}