00:01
So we have three related problems to work on, and the first one is the tridium, and you want to find what the half -life is, and we know that the decay rate is set up where it is 5 .5 % decaying per year, per year, which is having a k value of k value of negative 0 .055.
00:26
And so not seeing what your previous questions look like, i'm going to set it up to interpret the way i'm, and then you can adjust that if need be.
00:38
And that means that our multiplier here for time, we start with some initial amount, and that our multiplier with time is this, or our initial amount, and then one minus 0 .055 gives us 0 .94 here.
00:53
That's our multiplier for yearly decay.
00:57
Now, if you were using this interpreting it as initial amount of e to the and using the negative 0 .055t, then i'll show you what you do slightly differently.
01:09
So whether this is the compound a continuous rate or just the annual rate, that's what i don't know from your problem.
01:17
So i'm going to assume it is this.
01:19
And we want that to equal whatever the initial amount is times 1 1 .5 to the t over whatever the half life is.
01:30
And i'll just call h capital h the half life.
01:33
So we want these two models to be equal.
01:36
So we know that we want one half.
01:39
They both have the power of t and they both have multiplied to the i to the initial amount.
01:44
So we want half to the 1 over h to equal.
01:49
Is 0 .945.
01:51
That's what we need to find is that h.
01:54
So how do we solve this problem? well, we can take the log of both sides, and that removes our exponent down, and we have 1 over h times the natural log of 1 half, which i'll just call 0 .5, since the others as a decimal.
02:13
And then we need to solve for h.
02:15
So i'm going to multiply both sides by h, and then take natural log of 0 .5, and divide divided by natural log of .945...