MATH SOLVE

4 months ago

Q:
# one hundred grams of radium are stored in a container. The amount R of radium present after t years can modeled by R=10e^-0.00043t. After how many years will only 5 grams of radium be present? round your answer to the nearest whole year.

Accepted Solution

A:

Using this equation, we get that it would take 1612 years for only 5 grams to be present.

Substituting 5 for R, we have

[tex]5=10e^{-0.00043t}[/tex]

Dividing both sides by 10,

[tex]\frac{5}{10}=\frac{10e^{-0.00043t}}{10} \\ \\0.5=e^{-0.00043t}[/tex]

Taking the natural log of both sides, we have

[tex]\ln{0.5}=\ln{e^{-0.00043t}} \\ \\ \ln{0.5}=-0.00043t[/tex]

Dividing both sides by -0.00043, we have

[tex]\frac{\ln{0.5}}{-0.00043}=t \\ \\1611.9=t \\ \\1612\approx t[/tex]

Substituting 5 for R, we have

[tex]5=10e^{-0.00043t}[/tex]

Dividing both sides by 10,

[tex]\frac{5}{10}=\frac{10e^{-0.00043t}}{10} \\ \\0.5=e^{-0.00043t}[/tex]

Taking the natural log of both sides, we have

[tex]\ln{0.5}=\ln{e^{-0.00043t}} \\ \\ \ln{0.5}=-0.00043t[/tex]

Dividing both sides by -0.00043, we have

[tex]\frac{\ln{0.5}}{-0.00043}=t \\ \\1611.9=t \\ \\1612\approx t[/tex]