Question

suppose ( w(t) ) denotes the amount of a radioactive material left after time ( t ). assume that ( w(0)=9 ) and ( w(1)=3 ). complete parts (a) through (c)
(a) find the differential equation that describes this situation.
the differential equation is ( \frac{dw}{dt}=(-ln 3)w(t) )
(b) how much material is left at time ( t = 5 )?
the amount of material left after ( t = 5 ) is ( w(5)=square ).
(type an exact answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find the general solution of the differential equation

The general solution of the differential equation \(\frac{dW}{dt}=- \ln 3W(t)\) is \(W(t)=W(0)e^{-\ln 3t}\). Since \(W(0) = 9\), then \(W(t)=9e^{-\ln 3t}\). Using the property \(a\ln b=\ln(b^{a})\) and \(e^{\ln x}=x\), we can rewrite \(W(t)\) as \(W(t)=9\times3^{-t}\) (because \(e^{-\ln 3t}=e^{\ln(3^{-t})}=3^{-t}\)).

Step2: Calculate \(W(5)\)

Substitute \(t = 5\) into the formula \(W(t)=9\times3^{-t}\). Then \(W(5)=9\times3^{-5}\). Using the rule \(a^{-n}=\frac{1}{a^{n}}\), we have \(W(5)=\frac{9}{3^{5}}\). Since \(3^{5}=243\), then \(W(5)=\frac{9}{243}=\frac{1}{27}\).

Answer:

\(\frac{1}{27}\)