Question

A scientist is measuring the amount of radioactive material in an unknown substance. When he begins measuring, there are 6.2 grams of radioactive substance. 12 days later, there are 6.16 grams. After 19 days, there are 6.13 grams. After 46 days, there are 5.92 grams. Assuming that the decay is exponential, find the equation that determines the number of grams remaining after x days.
A. y = 6(0.92)x
B. y = 6.15(0.98)x
C. y = 6.23(0.999)x
D. y = 6(0.999)x

I got A as my answer, is that correct?

Answer 1

@Directrix

Answer 2

was my answer correct?

Answer 3

@LolWolf im not sure if your typing, or if your computer froze up

Answer 4

Not quite.

We have a system to solve. Say we have \(M\) mass at time \(t\), then:
\[
\begin{align}
M(0)=6.2=a(b)^t&=a(b)^0=a\\
M(12)=6.16&=6.2(b)^{12}\\
\end{align}
\]Taking the \(\log\) on both sides to rid of the exponent, we have:
\[
\begin{align}
6.16&=6.2(b)^{12}\implies\\
\frac{6.16}{6.2}&=b^{12}\implies\\
\ln\left(\frac{6.16}{6.2}\right)&=12\ln(b)\implies\\
-.00647251&=12\ln(b)\implies\\
\ln(b)&=\frac{-.00647251}{12}\implies\\
b&=e^{\frac{-.00647251}{12}}\implies\\
\end{align}
\]So, we have:
\[
b=.999461
\]As our final result.

Answer 5

Sorry, I was helping another while typing this question.

Answer 6

So C would then be our answer correct? and its fine:)

Answer 7

?

Answer 8

Is there an error on answer C? Or in the question?

Checking with the last value, I get:
\[
6.2(.999)^{46}\approx 5.92\text{ grams}
\]Which is what's given.

Answer 9

no, no error.

Answer 10

maybe you have to round up some numbers, im not really sure.

Answer 11

Huh. Perhaps. But, yes I believe that C is the answer.

Also, how did you originally get answer A?

Answer 12

Someone had helped me through that problem awhile ago, and i just wanted to make sure it wa correct, but i guess not lol

Answer 13

@katlin95 I agree with @LolWolf on this.