The growth or decay of an original quantity C that increases or decreases in a p% per year after t years is given by the following equation:
[tex]p(t)=C\cdot(1\pm\frac{p}{100})^t[/tex]If the quantity increases (i.e. it growths) we use the + symbol inside the parenthesis. If the quantity decreases we use the - symbol. This implies that for a growth the term that is raised to t is greater than 1 and for a decay that term is smaller than 1.
Now let's compare that generic equation with the function given by the question:
[tex]3000\cdot(1.019)^t=C\cdot(1\pm\frac{p}{100})^t[/tex]One of the first things you can notice is that C=3000 which means that the initial price was $3000. Just to be sure that this is correct we can evaluate p(t) at t=0:
[tex]p(0)=3000\cdot(1.019)^0=3000[/tex]So the initial price was $3000.
Now let's compare the terms inside parenthesis that are raised to t:
[tex]1.019=1\pm\frac{p}{100}[/tex]As I stated before, if the term raised to t is greater than 1 then we are talking about a growth. 1.019 is greater than 1 so this function represents a growth. What's more, in the right side of the equation we must use the + symbol. This way we have an equation for the yearly percentage of change of the price:
[tex]1.019=1+\frac{p}{100}[/tex]We can substract 1 from both sides of this equation:
[tex]\begin{gathered} 1.019-1=1+\frac{p}{100}-1 \\ 0.019=\frac{p}{100} \end{gathered}[/tex]And we multiply both sides by 100:
[tex]\begin{gathered} 100\cdot0.019=\frac{p}{100}\cdot100 \\ 1.9=p \end{gathered}[/tex]So each year the price increases in a 1.9%.
AnswerThen the answers in order are:
$3000
growth
1.9%
