Question

I need to make a Half-Life function out of this.

F(x) = 37.23e^-0.55x
y = 37.23e^-0.55x
y/37.23 = e^(-0.55x)
ln y/37.23 = ln e^(-0.55x)
ln y – ln 37.23 = ln e^(-0.55x)
(3.62)
ln y = ln e^(-0.55x) + 3.62
ln y = -0.55x + 3.62

Answer 1

@satellite73 hopefully you can help me.

Answer 2

@hba, I am still looking for assistance! D:

Answer 3

\[F(x) = 37.23e^{-0.55x}\]

Answer 4

set
\[.5=e^{-.55x}\] and solve for \(x\) to find the half life

Answer 5

take two steps
1) take the log
2) divide by \(-.55\)

Answer 6

\[e^{-.55x}=.5\]
\[-.55x=\ln(.5)\]
\[x=\frac{\ln(.5)}{-.55}\] then a calculator

Answer 7

So I have to solve for X and fill in the blanks you left for me?

Answer 8

half life is the time it takes to get half of what you started with
the coefficient in front of the \(e\) is unimportant, since half of it will be, well, half
in general if you have
\[Pe^{kt}\] and you want the half life, half of \(P\) is \(\frac{P}{2}\) so you could start with
\[Pe^{kt}=\frac{P}{2}\] but the first step is to divide by \(P\) and you get
\[e^{kt}=\frac{1}{2}=.5\] so you might as well start with
\[e^{kt}=.5\]
go to
\[kt=\ln(.5)\] and so
\[t=\frac{\ln(.5)}{k}\]

Answer 9

I am so lost, but I do appreciate the help haha.

Answer 10

the problem with your solution above is that you kept the variable \(y\) but for the half life, \(y\) is not a variable, it is known to you
namely, it is half of what you started with

Answer 11

lets start from scratch because it is not too hard

Answer 12

So we start with F(x) = 37.23e^-0.55x?

Answer 13

you have
\[F(x) = 37.23e^{-0.55x}\] as your original decay function

Answer 14

now you want the half life
i.e. how long before you have half of what you start with

Answer 15

Couldn't I just divide it by 2, or do I need to have something else to go with it?

Answer 16

half of \(37.25\) is \(\frac{37.25}{2}\) whatever that is

Answer 17

so now we know what \(F(x)\) is, it is \(F(x)=\frac{37.25}{2}\) that is a number known to us

Answer 18

It comes out to 18.615

Answer 19

what we want to solve for is \(x\)
that is, we want to solve
\[\frac{37.25}{2}=37.25e^{-.55x}\]

Answer 20

By the way, its 37.23! You confused me for a second.

Answer 21

ok whatever, the point is that it does not matter what you start with

Answer 22

So 37.23/2 = 18.6 which is equal to 37.23e^-.55x?

Answer 23

because the first step in solving
\[\frac{37.23}{2}=37.23e^{-.55x}\] is to divide both sides by \(37.23\)

Answer 24

so the \(37.23\) is irrelevant to the problem, you should go right to
\[\frac{1}{2}=e^{-.55x}\]

Answer 25

make sure you understand that the initial amount, the number in front of the \(e\) is not important to the problem
whatever you have there, half of it is half of it, so when you divide both sides by that number, you will get \(\frac{1}{2}\) no matter what number you start with

Answer 26

if you have
\[100e^{-.55x}\] then you would say half of 100 is 50 and get
\[100e^{-.55x}=50\] which still comes down to solving
\[e^{-.55x}=\frac{1}{2}\]

Answer 27

Okay, so 1/2 = e^-.55x

Answer 28

right
now you are done in two steps
take the log get
\[\ln(.5)=-.55x\] and finally
\[x=\frac{\ln(.5)}{-.55}\]
it always works this way
takes two steps only

Answer 29

-1.243147181?

Answer 30

the half life should be positive
this is what i get
http://www.wolframalpha.com/input/?i=-ln%28.5%29%2F.55

Answer 31

Never knew about that site thanks for that, I think I just put it in incorrectly on my calculator xD

Answer 32

wolf is the best way to cheat i have found
tells you everything

Answer 33

of course you have to know what to ask it
here i solved directly without taking the log
http://www.wolframalpha.com/input/?i=.5%3De^%28-.55x%29

Answer 34

So that would be the final answer for that part?

Answer 35

Once again @satellite73 I really appreciate the help thank you.