Twinkies on the shelf of a convenience store lose their fresh tastiness over time. We say that the taste quality is 1 when the Twinkies are first put on the shelf at the store, and that the quality of tastiness declines according to the function [tex]$Q(t)=(0.85)^t$[/tex], where [tex]$t$[/tex] is measured in days. Determine when the taste quality will be one half of its original value. Round to 2 decimal places.

The taste quality will be one half its original value after [tex]$\square$[/tex] days.



Answer :

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.