Respuesta :
We are given
Scientist begins with 150 milligrams of a radioactive substance
so, [tex]P_0=150[/tex]
we can use formula
[tex]P(t)=P_0e^{kt}[/tex]
we can plug Po=150
[tex]P(t)=150e^{kt}[/tex]
After 29 hours, 75 mg of the substance remains
so, at t=29 , P=75
we can plug it
[tex]75=150e^{k\times 29}[/tex]
now, we can solve for k
[tex]e^{k29}=\frac{1}{2}[/tex]
now, we can take ln on both sides
[tex]\ln \left(e^{k\times \:29}\right)=\ln \left(\frac{1}{2}\right)[/tex]
[tex]k=-\frac{\ln \left(2\right)}{29}[/tex]
[tex]k=-0.02390[/tex]
now, we can plug it back
and we get
[tex]P(t)=150e^{-0.02390t}[/tex]
we can plug t=33
[tex]P(33)=150e^{-0.02390\times 33}[/tex]
we get
[tex]P(33)=68.16528mg[/tex]..............Answer
Answer:
The required weight after [tex]33[/tex] hours is [tex]68.162\;\rm{mg[/tex].
Step-by-step explanation:
Given: A radioactive substance decays exponentially. A scientist begins with [tex]150[/tex] milligrams of a radioactive substance. After [tex]29[/tex] hours, [tex]75[/tex] mg of the substance remains.
As per given information:
A Scientist begins with [tex]150[/tex] milligrams of a radioactive substance,
Let us take [tex]P_o=150[/tex]
Using the formula: [tex]P(t)=P_oe^kt[/tex]
Substiuting all the values as [tex]t=29, \; P=75 \;\&\;P_o=150[/tex]
[tex]\;\;\;75=150e^{29k}\\e^{29k}=\frac{1}2[/tex]
Now, taking [tex]log[/tex] both side:
[tex]29ln(e^k)=ln(\frac{1}{2})\\[/tex]
[tex]k=\frac{-ln2}{29}\\k=-0.02390[/tex]
Here, the equation becomes: [tex]P(t)=150e^{-0.02390t[/tex]
Then substituting the value of [tex]t=33:[/tex]
[tex]P(33)=150e^{-0.02390\times 33}\\P(33)=68.1652\;\rm{mg[/tex]
Hence, [tex]68.162\;\rm{mg[/tex] will remain after 33 hours.
Learn more about exponential function here:
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