Question

The radioactive substance strontium-90 has a half-life of 28 years. In other words, it takes 28 years for half of a given quantity of strontium-90 to decay to a non-radioactive substance. The amount of radioactive strontium-90 still present after t years is modeled by the expression grams. Evaluate the expression for t = 0, t = 28 and t = 56. What does each value of the expression 8(2^-t/28) represent?

Answer 1

Do you know the formula for radioactive decay?

Answer 2

no @geoffb

Answer 3

8(2^-t/28) @geoffb ?

Answer 4

Right. The question is asking what each value in that equation represents.

Answer 5

so how do i answer that?

Answer 6

Start by pulling out each value. 8 is a value. t is a value. 28 is a value. Et cetera.

Answer 7

What do each of those represent?

Answer 8

Uh, . . . that's not what's going on here, @geoffb

Answer 9

It's the last sentence in the question. I figured that would be a better place to start, since he didn't know the formula. From there, he can plug in values of t and determine Q.

Answer 10

" Evaluate the expression, 8(2^-t/28), for t = 0, t = 28 and t = 56."

The formula was given in the problem statement.

Answer 11

"What does each value ... represent?" means to interpret the output values after evaluating.

Answer 12

Ahhh, yes. That makes more sense

Answer 13

so i take this equation 8(2^-t/28) and change it to 8(2^-0/28), 8(2^-28/28), and 8(2^-56/28)??

Answer 14

I'd recommend a T-table like this:

Answer 15

Yes, @chicagochica5 exactly!

Answer 16

@CliffSedge and do i put them all negative since it was -t, or do i change it to positive?

Answer 17

Yes, the - makes those values negative before you divide them by 28.

Answer 18

@CliffSedge, for the past hour ive been working the problem and i dont know how to get the answers :( please help

Answer 19

\[A(t)=8(2^{-t/28})\]
First find A(0), A(28), and A(56)
\[A(0)=8(2^{-(0)/28})\] \[A(28)=8(2^{-(28)/28})\] \[A(56)=8(2^{-(56)/28})\]

Answer 20

@chicagochica5 were you able to find those 3 amounts yet?

Answer 21

@CliffSedge no i cant solve them

Answer 22

Simplify the exponents:
\[A(0)=8(2^0)\]\[A(28)=8(2^{−1})\]\[A(56)=8(2^{−2})\]
Does that help?

Answer 23

the first one, No solution?
the next one \[A = \frac{ 2 }{ 7 }\] ?

Answer 24

and the last one \[A= \frac{ 1 }{ 28 }\] ? @CliffSedge

Answer 25

Can you try \[2^{1}\] and \[2^{2}\]

Answer 26

where

Answer 27

A negative exponent is the reciprocal of the positive exponent. So, if you solve 2^1 and 2^2, take the reciprocal (one over the result) to get 2^-1 and 2^-2.

Answer 28

Anything (except zero) to the 0th power = 1.
Negative exponents mean to divide by that power.

Answer 29

I'll just let my buddy, geoffb take it from here. ;-)

Answer 30

im confused

Answer 31

Okay sorry. I'll try again.

Answer 32

its not you, im not good at math

Answer 33

Do you know how to solve a number "to the power of" something, when it is positive?

For example, do you know how to do 2 to the power of 2?
\[2^{2}\]

Answer 34

is 4?

Answer 35

Yes, it is, but how did you figure that out? What exactly did you do?

Answer 36

2 x 2

Answer 37

Exactly. So what would be:

\[2^{3}\]?

Answer 38

And I don't need the answer—what would you do to *get* the answer?

Answer 39

2 x 2 x 2

Answer 40

Okay, good! You have a good grasp of positive exponents. Now...

Answer 41

Do you know what a reciprocal is?

Answer 42

a fraction flipped around

Answer 43

Yes, exactly! Good. Now, do you know that every number can be represented as a fraction? For example, what would be the number 6 as a fraction?

Answer 44

6/1?

Answer 45

Yes, great! Which means you can always take the reciprocal by putting 1 over the number, because all you're doing is "flipping" the fraction. Does that make sense?

Answer 46

yes

Answer 47

Awesome. Now, as Cliff was about to say, a negative exponent simply means you take the reciprocal. So, the following are exactly the same.

\[2^{-3} = \frac{1}{2^{3}}\]
Still with me?

Answer 48

wheres that equation from? how did you get that

Answer 49

It's not from anywhere. I just picked it out of a hat.

I could have said:

\[x^{-2} = \frac{1}{x^{2}}\]
or
\[6^{-8} = \frac{1}{6^{8}}\]

Answer 50

The point I'm trying to get across is that a negative exponent simply means you "flip the fraction" and find the reciprocal.

Answer 51

oh ok, so 2^-3 = 1 over 2^3?

Answer 52

Yes! Awesome.

Answer 53

Now, going back to what Cliff wrote before, can you figure out these?:

\[A(0)=8(2^{0})\]
\[A(28)=8(2^{−1})\]
\[A(56)=8(2^{−2})\]

Answer 54

no i dont know how to solve

Answer 55

Okay. You seemed to a minute ago. Be confident in yourself! :)

Answer 56

If you had to take a wild guess, what would it be?

Answer 57

for the first one, i had No solution

Answer 58

The first one is actually, as Cliff said, any number (except 0) to the power of 0 is 1.

Answer 59

thats the answer?

Answer 60

Let's cut to the chase, so we can get to interpreting the values.

2^0 = 1
2^-1 = 1/2
2^-2 = 1/4

Answer 61

ok and those numbers mean

Answer 62

Multiplying those by the 8 grams you started with yields how much of the strontium is left after that amount of time.

Answer 63

multiply 0 x 8?

Answer 64

No, t is the number of years elapsed. When t = 0, you have 8 grams.