y=ab^x
In this equation a is the initial value and b is the multiplier. We know that it weighs 2 grams from the start, which means the function's initial value is a= 2.
y= 2b^x
Also, between 10:00a.m. and 6:00p.m. there is a difference of 8 hours. In this time the material decays to 0.45 grams. We can interpret this as the data point (8,0.45). By substituting this into the function we can find the value of b.
The multiplier is 0.83. We can substitute it for b to complete the equation modeling this situation.
y = 2b^x
⇔
y = 2( 0.83)^x
b In Part A we found that the approximate hourly multiplier is 0.83.
To find the hourly percent of decrease, we have to subtract this number from 1 and write the difference in percent.
1-0.83=0.17=17 %
The percent of decrease is 17 %.
c In Part A we found the following equation to model our situation.
y=2(0.83)^x
Here, 2 represents the initial value — the weight of the material at 10:00a.m.. We want to find what the weight was of this material at 8:00a.m., which is 2 hours before what we defined as the initial value. For this reason we have substitute x = -2 into the equation above to get the weight.
