Question

the quantity of a radioactive substance decays according to the function $q(t) = 100e^{-\frac{t}{4}}$, where $t$ represents time in years. choose all of the expressions equivalent to $q(t)$. a. $q(t) = -25e^t$ b. $q(t) = 25e^{-t}$ c. $q(t) = 100\left(e^{-2}\
ight)^{\frac{t}{8}}$ d. $q(t) = 100\left(e^{-\frac{1}{4}}\
ight)^t$ e. $q(t) = 100\left(-\frac{e}{4}\
ight)^t$ f. $q(t) = 400e^{-t}$

Explanation:

Step1: Use exponent rule \(a^{bc}=(a^b)^c\)

For \(Q(t) = 100e^{-\frac{t}{4}}\), we can rewrite it as \(Q(t)=100(e^{-\frac{1}{4}})^t\) (by the rule \(a^{bc}=(a^b)^c\) where \(a = e\), \(b=-\frac{1}{4}\), \(c = t\)).

Step2: Check each option

• Option A: \(Q(t)=-25e^{t}\). Since \(100e^{-\frac{t}{4}}

eq - 25e^{t}\) (coefficient and exponent sign are wrong).

• Option B: \(Q(t)=25e^{-t}\). \(100e^{-\frac{t}{4}}

eq25e^{-t}\) (coefficient and exponent are wrong).

• Option C: \(100(e^{-2})^{\frac{t}{8}}=100e^{-\frac{2t}{8}}=100e^{-\frac{t}{4}}\) (using \(a^{bc}=(a^b)^c\) with \(a = e\), \(b=-2\), \(c=\frac{t}{8}\)).
• Option D: As shown in Step 1, \(Q(t)=100(e^{-\frac{1}{4}})^t\) is correct.
• Option E: \(Q(t)=100(-\frac{e}{4})^t\). Since the base of an exponential function \(y = a^t\) for exponential growth/decay should be \(a>0,a

eq1\), and \(-\frac{e}{4}<0\), this is wrong.

• Option F: \(Q(t)=400e^{-t}\). \(100e^{-\frac{t}{4}}

eq400e^{-t}\) (coefficient and exponent are wrong).

Answer:

C. \(Q(t)=100(e^{-2})^{\frac{t}{8}}\), D. \(Q(t)=100(e^{-\frac{1}{4}})^t\)