How do I write a half life function of this info?
F(x)=45e^-.89x
y=45e^-.89x
y/45 = e^(-.89x)
in y/45 = In e^(-.89x)
in y - in 45 = in e^-.89x
in y = in e^-.89x+3.81
in y = -.89x+3.81

- anonymous

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- anonymous

HELP! It says it should look like this: F(x)=A(1/2)^x. Where A is the initial amount of the substance and x represents the time for decay. I have A, it´s 25. How do i find x?

- asnaseer

if:\[F(x)=45e^{-.89x}\]then the value of F(0) will tell you how much the initial amount is equal to. In this case it would be:\[F(0)=45*e^0=45\]

- asnaseer

what you need to find is the value of x to get half this value

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## More answers

- asnaseer

i.e. what value of x will give you F(x) = 22.5

- anonymous

So it would be, 45(1/2)^22.5 ?

- asnaseer

no, you need to solve the equation:\[22.5=45*e^{-0.89x}\]

- asnaseer

this will give you a value for x for which F(x) is equal to half its initial value

- anonymous

i plugged, 45*e^-.89 into my calculator and got 18.5. is that right?

- asnaseer

how does that solve the equation I listed above?

- asnaseer

you need to rearrange that equation to get an expression for x

- asnaseer

I can show you the first few steps...

- asnaseer

so, starting with:\[22.5=45*e^{-0.89x}\]we divide both sides by 45 to get:\[0.5=e^{-0.89x}\]then take logs of both sides to get:\[\ln(0.5)=-0.89x\]can you do the rest?

- anonymous

I´m lost sorry, no.

- asnaseer

which part are you stuck on?

- anonymous

what to do next

- asnaseer

do you understand that you need to find x?

- asnaseer

did you follow the reasoning above?

- anonymous

i understand i need to find x, but i dont understand how

- asnaseer

If I said:\[12=4x\]then would you be able to find the value of x here?

- anonymous

yes

- anonymous

3

- asnaseer

good, so now use the same principals to solve this for x:\[\ln(0.5)=-0.89x\]

- anonymous

i got -.56

- asnaseer

tat does not look right to me - please show me your steps so that I can check where you may have made a mistake

- asnaseer

*that

- anonymous

i divided .5 by -.89 or is it the other way around?

- asnaseer

you cannot do that - the left-hand-side of the equals sign has \(\ln(0.5)\) not just 0.5

- asnaseer

so first find the value for \(\ln(0.5)\), then divide that by -0.89

- anonymous

.78

- asnaseer

that looks more like it.

- anonymous

is that x?

- asnaseer

yes - you can double check that you have the right value for x by substituting it into your original equation:\[F(x)=45e^{-.89x}\]and see if you get roughly 22.5 as the answer.

- anonymous

why would my A be 45?

- anonymous

45(1/2)^.78

- asnaseer

In the function given to you, the value of the function when x=0 is 45.

- anonymous

is that the half life function?

- asnaseer

no

- anonymous

oh mah gaaahh. this is tuff

- asnaseer

In general you ay have an exponentially decaying function defined as:\[F(x)=Ae^{-bt}\]

- asnaseer

in this function t represents the time

- asnaseer

and you can see (I hope) that when t=0 F(0)=A

- asnaseer

so A represents the initial amount - i.e. the amount at the start

- asnaseer

as time passes, the mount decays and gets less and less

- asnaseer

*amount

- asnaseer

at some point, it decays to half its original value

- asnaseer

so, at that point F(t) = A/2

- asnaseer

now, since we know that:\[F(t)=Ae^{-bt}\]then we can replace F(t) by A/2 to get:\[\frac{A}{2}=Ae^{-bt}\]divide both sides by A to get:\[\frac{1}{2}=e^{-bt}\]take logs of both sides to get:\[\ln(0.5)=-bt\]divide both sides by -b to get:\[t=-\frac{\ln(0.5)}{b}\]this is what the half life is

- asnaseer

it represents the amount of time that has to elapse before the amount of material decays to half its original value

- anonymous

I understand the process but it tells me that it needs to look like this: A(1/2)^x

- asnaseer

I don't know where you are getting that from - it makes no sense to me.

- anonymous

The half-life of a substance is the time it takes for half of the substance to decay. The exponential function representing half-life is f(x) = A(1/2) where A is the initial amount of the substance and x represents the time for decay. The half-life of the substance will depend on the initial amount of substance you have. In this activity, you will experience this formula in action.

- anonymous

I had to use pennies in the beggining if that helps, so i thought that was A, i used 25 pennies.

- asnaseer

where you wrote:
f(x) = A(1/2)
that corresponds to what I said above when I said that at its half life:
so, at that point F(t) = A/2

- asnaseer

what this is saying is that when the material (whatever it may be) reaches its half life, then the value of the function f(x) will be equal to HALF the value of f(0).
and we know that f(0) = A
so, at its half life, f(x) = A/2

- anonymous

that´s the function? why doesnt it look like f(x) = A(1/2)^x ?

- asnaseer

it is saying that the VALUE of the function will equal A/2

- asnaseer

e.g. if I had a function defined as:\[f(x)=x^2\]and I asked you to find the value of x for which f(x) = 16, then you would do:\[16=x^2\]therefore:\[x=\sqrt{16}=4\]

- asnaseer

so note here that I said you need to find the value of x for which:
f(x) = 16

- asnaseer

so here we are saying that the VALUE of f(x) should be equal to 16

- asnaseer

regardless of how the function itself is defined

- anonymous

So how will the half life equation of my probelm look? like this f(x) = A/2 ?

- asnaseer

the HALF LIFE is defined as the point at which the VALUE of the function is equal to half its initial value - i.e. the point at which f(x) = A/2

- asnaseer

this is not stating that the FUNCTION definition is f(x) = A/2
instead it is stating that the VALUE of f(x) = A/2
and you need to find a suitable x for this to be true

- asnaseer

I need to go now - it is very late here and I need some sleep - hope you understand this concept better now.

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