Question

The function A=A_oe^{-0.01155x} A = A o e − 0.01155 x models the amount in pounds of a particular radioactive material stored in a concrete vault, where A_o A o is the original amount of material and x is the number of years since the material was put in the vault. If 900 pounds of material are placed in the vault, how much time will need to pass for only 357 pounds to remain? Round your answer to the nearest tenth.

Answer 1

They are essentially asking us to solve:

$357=900e^{-0.01155x} \implies\\ \frac{357}{900}=e^{-0.01155x} \implies\\ \ln{\frac{357}{900}}=-0.01155x \implies \\ x\approx 80.1$

Solving this equation for x tells us that it will take around 80.1 years for the materials in the vault to reduce to 357 pounds.

Answer 2

Answer:

80.1 years of time will be needed to pass.

Step-by-step explanation:

Given expression:

$A=A_o\times e^{-0.01155\times x}$

Given :$A_o=900 pound$

A = 357 pounds

x = ?

Using the given expression:

$A=A_o\times e^{-0.01155\times x}$

Taking ln both sides:

$\ln A=\ln A_o -0.01155\times x$

$\ln (357 pounds) =\ln (900) -0.01155\times x$

$\ln (357 pounds) -\ln (900 pounds)=-0.01155\times x$

$x=\ln \frac{900 pounds}{357 pounds}\times \frac{1}{0.01155}$

x = 80.0570 years ≈ 80.1 years

80.1 years of time will be needed to pass.