The mass of a substance, which follows a continuous exponential growth model, is being studied in a lab. The doubling time for this substance was observed to be 7 hours. There were 23.3 mg of the substance present at the beginning of the study.

(a) Let t be the time (in hours) since the beginning of the study, and let y be the amount of the substance at time t. Write a formula relating y to t. Use exact expressions to fill in the missing parts of the formula Do not use approximations.
(b) How much will be present in 5 hours?

Respuesta :

Answer:

[tex]a.\ \ \ y=23.3e^{0.09902t}\\\\b.\ \ \ 38.23\ mg[/tex]

Step-by-step explanation:

-This is an exponential relationship where one of the variables contains an exponent component.

The general form of an exponential function is given as:

[tex]A_n=A_oe^{rt}[/tex]

where:

[tex]r-rate \ of \ growth/decay\\t-time\\A_o-Initial \ Population\\A_{t/n}-Population \ at \ time\ t/n[/tex]

#Given the initial population as 23.3mg, doubling time as 7hrs, we determine the growth rate as:

[tex]y=23.3e^{7r}\\\\y=23.3\times 2=46.6\\\\\therefore 2=e^{7r}\\\\7r=In \ 2\\\\r=0.09902[/tex]

#Substitute r in the general equation for solving for amount at time t:

[tex]y=23.3e^{0.09902t}[/tex]

b. We use the amount at time t from a above,[tex]y=23.3e^{0.09902t}[/tex], to calculate the amount after 5hrs as:

[tex]y=23.3e^{0.09902t}\\\\=23.3^{0.09902\times 5}\\\\=38.23\ mg[/tex]

Hence, the mass of the substance after 5hrs is 38.23 mg

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