Question

Hey guys please help? Medal award :D
A sample of radioactive material has been decaying for 5 years. Three years ago (at 2 years ) there were 6.0grams of material left. Now (at 5 years) there are 5.2 grams of material left

a. Write two points from this problem
b. using the point-ratio y=y1b^x-x1 plug in the two points and solve for the ratio b.
c.write an exponential equation in point-ratio form for this situation

Answer 1

for the two points i believe it is (2,6.0) and (5,5.2)
for the point ratio i thnk it's 5.2=6.0b^5-2
c i'm not sure

Answer 2

Can someone please help..:(

Answer 3

i believe ur first two are correct but C i have no clue

Answer 4

@crabby_patty I'm not sure what they mean by exponential point ratio maybe y=ab^x?

Answer 5

since we have the two points what exactly are they looking for?

Answer 6

(2, 6.0) and (5,5.2)

Answer 7

find the common ratio, look at the ratios of consective y-values

Answer 8

5.2/6

Answer 9

and 5/2? or just 5.2/6 i'm just not aware of this concept

Answer 10

if you have more than one ratio then you obtain the mean

Answer 11

The point of this problem is to come up with an EXPONENTIAL model for the relationship between x and y. If we succeed in doing that, then either of the given points will "satisfy" the equation for this model.

Answer 12

your ratio determines the exponential decay

Answer 13

I had some trouble with "y=y1b^x-x1 at first. It took me a while to realize that there should be parentheses around " x - x1 ":\[y=y _{1}b ^{(x-x _{1})}\]

Answer 14

so would b and c be the same type of equation?

Answer 15

In other words, the quantity inside parentheses has to be calculated before it's used as the exponent of base b.

Answer 16

sorry I misspoke
the ratio would be multiplied to the first term
30* (ratio)^x

Answer 17

@sana1234 " That's right. One and the same equation applies to both Parts a and b.

Answer 18

@mathmale it's 5.2=6.0b^3?

Answer 19

you start with 6

Answer 20

I'd prefer to see how you obtained your results.
Have you checked both of the given points in your equation 5.2=6.0b^3 ?

Answer 21

@nincompoop 6=5.2b^3?

Answer 22

why cube?

Answer 23

y = 6(ratio)^x then you look at the proceeding points

Answer 24

@nincompoop for c? I was just going over my equation for b

Answer 25

x-x1 would be applied to the second point you have
remember that you are not starting from zero year here. you started with 2nd year
so it can mean that on the first year, the gram must be greater than the second year

Answer 26

@sana1234 I could help you more if you'd share your work. If you truly want to understand how to obtain an exponential model for this situation, you'll need to calculate a and b by substituting the given points {(3,6), (5.5.2)} into the model I gave you,\[y=y _{1}b ^{(x-x _{1})}\]

Answer 27

your y1 = 6 not 5.2

Answer 28

Have you actually done that?

Answer 29

@mathmale i'm just really confused haha just for clarification for b the equation is 5.2=6.0b^3 and why would it be (3,6) ?

Answer 30

Because you have two unknown quantities, a and b, and it is your job to find their values. To do that, you take the model given, and then for each point you substitute the x- and y-coordinates into the model. You'll end up with two equations and that is enough info to enable you to calculate a and b.

Answer 31

going back to what you've just said:
the equation is 5.2=6.0b^3 and why would it be (3,6) ?

Again, you have two separate points, (3,6) and (5, 5.2).
You have ONE model, which I've typed in before.

In each case, the point will give you x1 and y1. Substitute those values into the model. Our goal is to find a and b.

Answer 32

@mathmale it's (2, 6)

Answer 33

decaying for five years. then 3 years ago
it means on 2nd year of decay

Answer 34

What's (2,6)? (2,6) is a point. Was the objective of this problem to find a point?

Answer 35

@mathmale it's point ratio form

Answer 36

I understand we're supposed to use that particular form. While I have the background to figure it out, I haven't actually used that form myself. Can either of you find an example in your textbook (if you have one) or online? I learn by example. If you have an example, could you share an image of it with me?

There are other exponential models that would apply in this situation; were I able to find an expo model for the 2 given data points, then I could most probably convert that to the point-ratio form.

Answer 37

just a sec.. I believe i do

Answer 38

ok i have a really good example

Answer 39

An exponential decay function in point-ratio form y=y1*b^x-x1 has the horizontal asymptote y=o. because these data approach a long-run value of 1.25, the exponential function must be translated up 1.25 units. To do so replace y with y-1.25. THe coefficient y1 is also a y value, so you must also replace y1 with y1-1.25 in order to account for the translation

Answer 40

http://math.kendallhunt.com/documents/daa2/cl/daa2cl010_05.pdf

Answer 41

The point-ratio equation is now y-125=(y1-1.25)*b^x-x1. TO find the value of b, select one point for x1 y1 and a second point (not too close) for (x,y). If you choose (10,1.97) and (50,1.36) you will find 1.36-1.25=(1.97-1.25)*b^50-10
-substitute (10,97( for x1 y1 and 50, 1.36 for x,y

Answer 42

0.11-0.72b^40 subtract
b^40=0.11/0.72 "divide both sides by 0.72
b=(0.11/0.72)^1/40 is approximately 0.9541 (raise both sides to the power of 1/40

Answer 43

Wow, @sana1234, this is really good stuff, some of which is new to me. I have a suggestion: Could you possibly work on something else for a while and let me look over your course material so that I could come to a better understanding of it? Don't want to waste your time while I learn. But, on the other hand, you seem to be making great progress on your own, for which I'm thankful. What do YOU want to do now? Split, and re-convene later, or you go on your own path?

Answer 44

thanks so much :) i'll be on here i'm actually looking through some other examples haha i have roughly an hour and a half before i have to head to school though again thanks so much!

Answer 45

OK! I'll be working on this.

Answer 46

btw for y-1.25=(y1-1.25) *b^x-x1 I forgot to put a point for 1.25

Answer 47

@mathmale thanks again so much I really appreciate it a lot!

Answer 48

I'm still learning my way through this, but believe I have an approach that results in my obtaining the common ratio, b. I've obtained the same value for b in several different ways, which increases my optimism that my b is correct.

I start with the point-ratio formula\[y=y _{1}*b ^{(x-x _{1})}\]
and rewrite it, first as \[y=y _{1}b ^{x}b ^{-x _{1}}\]
and then as\[y=\frac{ y _{1} }{ b ^{x _{1}} }b^x\]

Answer 49

Makes sense :D

Answer 50

Our goal is to find b. Note that the coefficient of b^x, above, is just a constant.
Therefore, this equation could be modeled by \[y=ab^x\]
We have 2 points on the curve: (3,6) and (5,5.2).
Substituting them into the above equation,
\[6=ab^3 \] and\[5.2=ab^5\]

Solving these 2 equations simultaneously for b results in \[b^2=\frac{ 5.2 }{ 6 }=\frac{ 52 }{ 60 }=\frac{ 13 }{ 15 }\]
So, if my assumptions and math are correct,
\[b=\pm \sqrt{\frac{ 13 }{ 15}}\]

Hope this helps you to finish your posted problem.

Note, again, that there are multiple ways in which to solve this problem. If it happens that you and arrive at the same value for b using different methods, that assures us that we're on right (if different) tracks. Good luck!!

Answer 51

I believe it' correct :)) it makes sense.. so basically

Answer 52

@mathmale thank you so much for your time… I honestly really really do appreciate it :)

Answer 53

My great pleasure! From what I've seen of your work, you know a lot about this kind of math yourself.