# Math Help - Logistic function story problem

1. ## Logistic function story problem

"Suppose that the water in a hot water heater is 70° Celsius above room temperature. When you first turn on the faucet, the water coming out is only 2° above room temperature. Ten seconds later it has warmed to 15° above room temperature.

The general equation for a logistic function is:

$T(x)=c/(1+ae^-bx)$

Use the three given values to find algebraically the particular equation of this function.

I spent some time and worked it out:

t(x)=70/(1+ae^(-bx))

10=70/(1+ae^(-b*15))
10+(1+ae^(-b*15))=70
10+10ae^(-b*15)=70
10ae^(-b*15)=60
a=60/(10e^(-b*15))

a(10e^(-b*15))=60
10ae^(-b*15)=60
ae^(-15b)=6
log(ae^(-15b))=log(6)
log(a)-15b*log(e)=log(6)
-15b*log(e)=-log(a)+log(6)
b=(log(a)-log(6))/(15log(e))
b=(log(60/(10e^(-b*15)))-log(6))/(15log(e))
b=(log(60/(10e^(-b*15)))-0.7781512504)/(6.514417229)
b=(log(6(1e^15b))-0.7781512504)/(6.514417229)
b=(log(16.30969097^15b)-0.7781512504)/(6.514417229)
6.514417229b=log(16.30969097^15b)-0.7781512504
7.292568479=log(16.30969097^15b)
10^7.292568479=16.30969097^15b
19614104.26=16.30969097^15b
log(base 16.30969097)19614104.26=15b
6.01475867=15b
b=0.4009839113

a=60/(10e^(-b*15))
a=60/(10e^(-0.4009839113*15))
a=60/10e^-6.01475867
a=2456.562119

t(x)=70/(1+2456.562119e^(-0.4009839113x))

Unfortunately, when I tested a point, it was wrong.

What did I do incorrectly?

2. Originally Posted by wiseguy
"Suppose that the water in a hot water heater is 70° Celsius above room temperature. When you first turn on the faucet, the water coming out is only 2° above room temperature. Ten seconds later it has warmed to 15° above room temperature.

The general equation for a logistic function is:

$T(x)=c/(1+ae^-bx)$

Use the three given values to find algebraically the particular equation of this function.

I spent some time and worked it out:

t(x)=70/(1+ae^(-bx))
Okay. Any time you cite a formula you should specify what all the variables represent but I think I am with you up to here. I think that "T(x)" is the temperature (above room temperature) after the tap has been open x seconds.

10=70/(1+ae^(-b*15))
Now you have lost me! Are you saying that "T" is the time and "x" is the temperature? I recommend you verify that. If so, you are going to have the temperature going to infinity as the time goes to some constant rather than the other way around! I think you have "x" and "T" reversed.

10+10ae^(-b*15)=70
10ae^(-b*15)=60
a=60/(10e^(-b*15))

a(10e^(-b*15))=60
10ae^(-b*15)=60
ae^(-15b)=6
log(ae^(-15b))=log(6)
log(a)-15b*log(e)=log(6)
-15b*log(e)=-log(a)+log(6)
b=(log(a)-log(6))/(15log(e))
b=(log(60/(10e^(-b*15)))-log(6))/(15log(e))
b=(log(60/(10e^(-b*15)))-0.7781512504)/(6.514417229)
b=(log(6(1e^15b))-0.7781512504)/(6.514417229)
b=(log(16.30969097^15b)-0.7781512504)/(6.514417229)
6.514417229b=log(16.30969097^15b)-0.7781512504
7.292568479=log(16.30969097^15b)
10^7.292568479=16.30969097^15b
19614104.26=16.30969097^15b
log(base 16.30969097)19614104.26=15b
6.01475867=15b
b=0.4009839113

a=60/(10e^(-b*15))
a=60/(10e^(-0.4009839113*15))
a=60/10e^-6.01475867
a=2456.562119

t(x)=70/(1+2456.562119e^(-0.4009839113x))

Unfortunately, when I tested a point, it was wrong.

What did I do incorrectly?