To solve the problem of determining how long it will be before the quantity [tex]\( Q(t) \)[/tex] of an element in a storage unit is less than 1.5 kg, given the function:
[tex]\[ Q(t) = 4 e^{-0.00938 t} \][/tex]
we need to find the time [tex]\( t \)[/tex] when [tex]\( Q(t) < 1.5 \)[/tex].
First, we set up the equation based on the given inequality:
[tex]\[ 4 e^{-0.00938 t} = 1.5 \][/tex]
Next, we solve for [tex]\( t \)[/tex]. To do this, follow these steps:
1. Divide both sides by 4 to isolate the exponential term:
[tex]\[ e^{-0.00938 t} = \frac{1.5}{4} \][/tex]
2. Simplify the fraction on the right-hand side:
[tex]\[ e^{-0.00938 t} = 0.375 \][/tex]
3. To solve for [tex]\( t \)[/tex], we take the natural logarithm (ln) of both sides:
[tex]\[ \ln(e^{-0.00938 t}) = \ln(0.375) \][/tex]
4. Utilize the property of logarithms [tex]\(\ln(e^x) = x\)[/tex], which yields:
[tex]\[ -0.00938 t = \ln(0.375) \][/tex]
5. Solve for [tex]\( t \)[/tex] by dividing both sides by -0.00938:
[tex]\[ t = \frac{\ln(0.375)}{-0.00938} \][/tex]
Calculating [tex]\( \frac{\ln(0.375)}{-0.00938} \)[/tex]:
[tex]\[ \ln(0.375) \approx -0.98083 \][/tex]
[tex]\[ t = \frac{-0.98083}{-0.00938} \approx 104.56601844474693 \][/tex]
Now, we round this value to the nearest year:
[tex]\[ t \approx 105 \][/tex]
Therefore, it will take approximately 105 years for the quantity of the element to be less than 1.5 kg.
So, the correct answer is:
b. 105 yr
