To determine how many days it will take for the bacteria count in the food product to reach [tex]\(4{,}000{,}000\)[/tex], we can use the given exponential growth equation for bacteria:
[tex]\[ f(t) = 500 e^{0.1 t} \][/tex]
We need to find the value of [tex]\( t \)[/tex] (time in days) when [tex]\( f(t) = 4{,}000{,}000 \)[/tex].
Start by setting [tex]\( f(t) \)[/tex] equal to [tex]\( 4{,}000{,}000 \)[/tex]:
[tex]\[ 4{,}000{,}000 = 500 e^{0.1 t} \][/tex]
Next, divide both sides of the equation by 500 to isolate the exponential term:
[tex]\[ \frac{4{,}000{,}000}{500} = e^{0.1 t} \][/tex]
Simplifying the left side:
[tex]\[ 8{,}000 = e^{0.1 t} \][/tex]
To solve for [tex]\( t \)[/tex], we need to take the natural logarithm (ln) of both sides. The natural logarithm (ln) will help us get rid of the exponential [tex]\( e \)[/tex]:
[tex]\[ \ln(8{,}000) = \ln(e^{0.1 t}) \][/tex]
Using the property of logarithms that [tex]\(\ln(e^x) = x\)[/tex], the equation becomes:
[tex]\[ \ln(8{,}000) = 0.1 t \][/tex]
Now, solve for [tex]\( t \)[/tex] by dividing both sides by [tex]\( 0.1 \)[/tex]:
[tex]\[ t = \frac{\ln(8{,}000)}{0.1} \][/tex]
Using the numerical computation, the natural logarithm of 8,000 is approximately [tex]\( \ln(8{,}000) \approx 8.987 \)[/tex]. Therefore:
[tex]\[ t \approx \frac{8.987}{0.1} \][/tex]
[tex]\[ t \approx 89.87 \][/tex]
Thus, it will take approximately [tex]\( 89.87 \)[/tex] days before the bacteria count reaches [tex]\( 4,000,000 \)[/tex] and the product becomes inedible.
