Answer:
10.4
Step-by-step explanation:
So an exponential equation with a decay can be expressed as: [tex]f(x)=a(1-r)^t[/tex] where r is the decay rate. In this case is decreasing at 4% each year, this means that the r is equal to 0.04 (divide the percent by 100 to convert to decimal). So the equation is going to be: [tex]f(x)=a(0.96)^t[/tex]. The reason this works is because whenever you decrease by x%, the new value is (1-x)%, which is expressed in the formula. But the reason the exponent is there is because after 1 year it's going to be 0.96a, but after two years, it's going to be 0.96 the previous year which is 0.96a, so after two years it's going to be 0.96(0.96a), and then the next year it's 0.96 of the second year, which is then 0.96(0.96(0.96a))... and so on. So the amount of 0.96 increases as t increases. the a is going to be the initial amount, since then t=0, 0.96^0 will be equal to 1, which makes the expression a(1), which is just a. So this gives you the complete equation: [tex]p(x)=13000(0.96)^t[/tex]
Now to find when it's equal to 8,500 you need to start by setting it equal to 8,500 so I'll work through each step
Original Equation:
[tex]p(x)=13000(0.96)^t[/tex]
Set the equation equal to 8,500
[tex]8500=13000(0.96)^t[/tex]
Divide both sides by 13,000
[tex]0.653846\approx 0.96^t[/tex]
Rewrite in logarithmic form
[tex]log_{0.96}0.653846=t[/tex]
If you're calculator allows you to enter the base when calculating log, but this expression into your calculator to calculate the value of t, otherwise use the change of base formula
[tex]\text{change of base formula: }log_ba = \frac{log(a)}{log(b)}[/tex]
[tex]t=\frac{log0.653846}{log0.96}\\[/tex]
[tex]t\approx\frac{-0.1845}{-0.0177}[/tex]
[tex]t\approx10.4[/tex]
