Calculating Half-Life Using Calculus: Deriving λ from y = y0e-kt

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Homework Statement


This is the equation
y = y0e-kt


Homework Equations



b) Show using your expression for λ that if at time t1 the amount is y1, then at time t1 + λ it will be y1/2, no matter what t1 is.

The Attempt at a Solution



y1=y0ekt1

that part I got. Then the answer key goes into the next step saying:λ = (−ln2/k)y0ek(t1+λ)

That is where I got lost. What is the step to go from the y1 equation into the lambda equation? I know that for half life, y1=y0/2 so that makes t=ln2/k . The time it takes for an element to decompose to half its mass is t=ln2/k . How do I make the connection to the very last lambda equation?
 
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The statement "at time t1 the amount is t1" is the same as y1=y0e-kt1. At any time, the amount is y=y0e-kt.
At time t=t1+λ, the amount is y=y0e-k(t1+λ)=y0e-kt1-kλ. You know the identity ax+y=ax ay, don't you? And λ is the half-life, so y0e-kλ=y0/2.
Can you proceed from here?


ehild
 
using your method and identity, I got: y=y0e-kt1*e-k[itex]\lambda[/itex]
so
y=y1*e-k[itex]\lambda[/itex]

and then I am stuck again. lambda = half life, but what does that mean for me to get to y0e-kλ=y0/2.
 
I still do not see it. is lambda in this equation being treated as a variable or constant? is lambda the same as time?
 
Ok I think I got it. This is my understanding, tell me if I am correct.

e-k[itex]\lambda[/itex] is the multiplier it takes to make an element exactly 1/2 of itself therefore we can set e-k[itex]\lambda[/itex]=1/2
So when I come to y=y0e-kt1*e-kλ That is the same thing as y=y0e-kt1*(1/2) and knowing that y0e-kt1=y1 we can say y=y1*(1/2)

am I right?
 
THank you all those who helped.