Question

A substance has a half-life of 46 years. What percent of the original amount of the substance will remain after 20 years? Round to the nearest percent.

Answer 1

44%

Answer 2

Can you please show me step by step how to do that?

Answer 3

Well, actually I am not quite sure but following my instincts I would think to find the percent, you would need to divide 20 by 46

20/46 = 0.43478260869...
I would round that so 0.44

So that's 44%

So if 44% is used up then 56% of the original substance is still available

Not sure if this is the right way to do it but it seems logical :)

Answer 4

I know I use the equation Po/2 = (Po)e^kt.

I found a similar problem that says: "A substance has a half-life of 56 years. What percent of the original amount of the substance will remain after 20 years? Round to the nearest percent." And the answer was 78%.
If I just do 20/56 I get 0.357, which would not be 78%.

I sense I'm not understanding the equation itself.
On the test I put the above equation, which the teacher said was right.
But when I do...

46/2 = 46e^k20, I was wrong.

Answer 5

This is all very confusing to me. What grade are you in? I'm in 9th

Answer 6

This is college pre-calculus.

Answer 7

Well that might explain why I don't understand it!

Answer 8

it is not right i don't think

Answer 9

try computing
\[\left(\frac{1}{2}\right)^{\frac{20}{46}}\]

Answer 10

maybe you get \(.44\) i don't know but it seems unlikely

Answer 11

=(1/2)^20/46 =0.5^0.43 =0.74

Let me try with the other sample problem to see if I get the right answer.

Answer 12

http://www.wolframalpha.com/input/?i=%281%2F2%29^%2820%2F46%29

Answer 13

(1/2)^20/56
=0.5^ 0.36
=0.78= 78%

Yeah, looks right. Now I need to think of it as an actual equation...

Answer 14

i used only the number given

Answer 15

half life is \(46\) so i used \(\left(\frac{1}{2}\right)^{\frac{t}{46}}\) for \(t=20\)

Answer 16

I think I'm having trouble imagining it as Po/2=Poe^kt, which I'm not going to get credit for unless I can relate it to that. Hmm...

Answer 17

we can do that too, it takes tons more work, but is not that hard

Answer 18

half life is \(46\) years which means you can set
\[e^{k\times 46}=\frac{1}{2}\] to solve for \(k\)

Answer 19

Okay, so Po is just treated as "1" since it's the initial year, right?

Answer 20

in equivalent logarithmic form you get
\[46x=\ln(\frac{1}{2})\] and so
\[k=\frac{\ln(.5)}{46}\]

Answer 21

the \(P_0\) makes no difference
if you start with 10 after 46 years you have 5
if you start with 1000 after 46 years you have 500
if you start with \(P_0\) after 46 years you have \(\frac{1}{2}P_0\)
so first step in solving
\[P_0e^{46k}=\frac{1}{2}P_0\] is to divide by \(P_0\) and start at
\[e^{46k}=\frac{1}{2}\]

Answer 22

notice it asks for "what percent remains" and doesn't tell you what you start with in any case
the initial amount is unimportant