Standardized Test Practice
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Use the data given in the table to calculate the rate of growth of the invested money.
Substitute A with 500 0000 into the function from Part A.
A(t)=250 000(1.035)^t
2018
Let's begin with recalling the exponential growth function.
A(t)=a(1+r)^t
Here, a is the initial amount, r is the rate of growth, and t is the time. It is given that Sandy inherited and then invested $250 000 in 1998. Hence, 250 000 is the initial value a.
Rearrange equation
.LHS /250 000.=.RHS /250 000.
sqrt(LHS)=sqrt(RHS)
Round to 3 decimal place(s)
LHS-1=RHS-1
Now that we know the value of r, we can substitute it into the function. A(t)=250 000(1+ 0.035)^t ⇓ A(t)=250 000(1.035)^t This equation can be used to predict how much money Sandy will have t years after investing.
Rearrange equation
.LHS /250 000.=.RHS /250 000.
sqrt(LHS)=sqrt(RHS)
Round to 3 decimal place(s)
LHS-1=RHS-1
As we can see, we got exactly the same value of rate of growth. Substituting it into the equation, we will get the function that predicts the amount of money Sandy will have t years after investing. A(t)=250 000(1.035)^t
To find in what year Sandy will have $500 000, we need to substitute A with 500 000 into the function from Part A.
500 000=250 000(1.035)^t
.LHS /250 000.=.RHS /250 000.
ln(LHS)=ln(RHS)
ln(a^b)= b*ln(a)
.LHS /ln2.=.RHS /ln2.
Rearrange equation
It will take about 20 years for the invested money to be worth $500 000. Adding 20 to 1998, the year Sandy invested, we can find in what year it will happen. 1998+20=2018