Certainly! Let's determine the day when the taste quality of the Twinkies will be half of its original value.
We start with the given quality function:
[tex]\[ Q(t) = (0.85)^t \][/tex]
where [tex]\( Q(t) \)[/tex] is the taste quality at day [tex]\( t \)[/tex].
We want to find the day [tex]\( t \)[/tex] when the taste quality is half of its original value, which means we set [tex]\( Q(t) \)[/tex] to [tex]\( 0.5 \)[/tex] (since half of 1 is 0.5).
So, we need to solve:
[tex]\[ (0.85)^t = 0.5 \][/tex]
To solve for [tex]\( t \)[/tex], we take the logarithm of both sides of the equation. Using logarithms:
[tex]\[ \log((0.85)^t) = \log(0.5) \][/tex]
By applying the power rule of logarithms ([tex]\( \log(a^b) = b \log(a) \)[/tex]), we get:
[tex]\[ t \log(0.85) = \log(0.5) \][/tex]
Next, solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\log(0.5)}{\log(0.85)} \][/tex]
Using this logarithmic expression and calculating the values:
- The logarithm of 0.5 is approximately -0.3010.
- The logarithm of 0.85 is approximately -0.0706.
So:
[tex]\[ t = \frac{-0.3010}{-0.0706} \approx 4.265 \][/tex]
Finally, rounding this result to two decimal places, we find:
[tex]\[ t \approx 4.27 \][/tex]
Thus, the taste quality will be one half its original value after [tex]\(\boxed{4.27}\)[/tex] days.
