How to find the inverse of a function

[Kupkake303]

T(t) = T_s+(98.6 – T_s)e^-kt

rewrite in the form t=g^-1(T)

In trying to understand how to find the inverse of this but am having a hard time, please advise.

Thanks,
Kupkake303

 

[stevendaryl]

This is just a matter of trying to get [itex]t[/itex] by itself on one side of an equality. You initially have: [itex]T = T_s + (98.6 - T_s) e^{-kt}[/itex]. To simplify the expression, let [itex]U = e^{-kt}[/itex]. Then the equation is: [itex]T = T_s + (98.6 - T_s) U[/itex]. So solve for [itex]U[/itex]. Then solve for [itex]t[/itex] in terms of this value of [itex]U[/itex].

 

[Kupkake303]

This is just a matter of trying to get [itex]t[/itex] by itself on one side of an equality. You initially have: [itex]T = T_s + (98.6 - T_s) e^{-kt}[/itex]. To simplify the expression, let [itex]U = e^{-kt}[/itex]. Then the equation is: [itex]T = T_s + (98.6 - T_s) U[/itex]. So solve for [itex]U[/itex]. Then solve for [itex]t[/itex] in terms of this value of [itex]U[/itex].

So I would put it as 0=T_s+(98.6-T_s)U ??

Doing that would give me -T_s/(98.6-T_s)=U

Then putting the e^-kt back in you get -T_s/(98.6-T_s)=e^-kt

So then I would then need to get that by it's self so the t is by itself but I can't remember how to do this. Please advise. Thank you

 

[stevendaryl]

So I would put it as 0=T_s+(98.6-T_s)U ??

No, [itex]T = T_s + (98.6 - T_s)U[/itex]

So [itex]U = (T-T_s)/(98.6 - T_s)[/itex]

 

[Kupkake303]

No, [itex]T = T_s + (98.6 - T_s)U[/itex]

So [itex]U = (T-T_s)/(98.6 - T_s)[/itex]

Right, because the T wouldn't just disappear.

So next I would change the U back to a e^-kt for e^-kt=(T-T_s)/(98.6-T_s)

from there would I try to figure out the e^-kt ?

 

[stevendaryl]

Right, because the T wouldn't just disappear.

So next I would change the U back to a e^-kt for e^-kt=(T-T_s)/(98.6-T_s)

from there would I try to figure out the e^-kt ?

You want to take the natural log of both sides of the equation.

 

[Kupkake303]

You want to take the natural log of both sides of the equation.

okay so I would
e^-kt=(T-T_s)/(98.6-T_s)

lne^-kt=ln(T-T_s)/(98.6-T_s)

The ln would cancel out the e

-kt=ln(T-T_s)/(98.6-T_s)

then divide by -k? to get

t=(1/-k)ln(T-T_s)/(98.6-T_s)...?

so this would be my answer for finding the inverse ..?

 

[stevendaryl]

okay so I would
e^-kt=(T-T_s)/(98.6-T_s)

lne^-kt=ln(T-T_s)/(98.6-T_s)

The ln would cancel out the e

-kt=ln(T-T_s)/(98.6-T_s)

then divide by -k? to get

t=(1/-k)ln(T-T_s)/(98.6-T_s)...?

so this would be my answer for finding the inverse ..?

Yes, except that [itex]-ln(A/B) = ln(B/A)[/itex], so you can get rid of the [itex]-[/itex] sign to get:

[itex]t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))[/itex]

 

[Kupkake303]

Yes, except that [itex]-ln(A/B) = ln(B/A)[/itex], so you can get rid of the [itex]-[/itex] sign to get:

[itex]t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))[/itex]

Awesome! Thank you soooo much. I spent about 4-5 hours trying to figure this problem out myself and I wouldn't have gotten through it by myself.

 

[epenguin]

Logarithms are the inverse of the finish of exponentiation, by definition. Hopefully you remember your first lesson about them. Well hopefully they told you that then.