Guided Example #1 Exponential Functions

The following are examples of exponential growth.

Example 1

If $3,000 is invested in an account at an annual rate of 6% compounded quarterly, how much money will be in the account after 5 years?

We are looking for A when P = $3,000, t = 5 years, r = .06, and n = 4.

$A=P\left(1+\frac{\:r}{n}\right)^{nt}\:\longrightarrow\:A=3,000\left(1+\frac{.06}{4}\right)^{\left(4\right)\left(5\right)}$ $A=P\left(1+\frac{\:r}{n}\right)^{nt}\:\longrightarrow\:A=3,000\left(1+\frac{.06}{4}\right)^{\left(4\right)\left(5\right)}$

$A = 3, 000 {(1.015)}^{20}$ $A=3,000\left(1.015\right)^{20}$

$A = 3, 000 (1.346855)$ $A=3,000\left(1.346855\right)$

$A=\$4,040.57$ $A=\$4,040.57$

Example 2

If $3,000 is invested at an annual rate of 6% interest compounded continuously, how much money will be in the account after 5 years?

We are looking for A using a continuous interest formula when P = 3,000, r = .06 and t = 5.

$A = P^{e r t}$ $A=Pe^{rt}$

$A = 3, 000 e^{(. 06) (5)}$ $A=3,000e^{\left(.06\right)\left(5\right)}$

$A = 3, 000 (1.3498588)$ $A=3,000\left(1.3498588\right)$

$A=\$4,049.58$ $A=\$4,049.58$

Example 3

If $3,000 is invested at an annual rate of 6% interest compounded continuously, how much money will be in the account after 5 years?

We are looking for A using a continuous interest formula when P = 3,000, r = .06 and t = 5.

If a car loses 25% of its value in one year, that means it is worth 75% of its initial value. At the end of the first year $=\$20,000\left(.75\right)\:=\:\$15,000$ $=\$20,000\left(.75\right)\:=\:\$15,000$ . At the end of the second year $=\$15,000\left(.75\right)\:=\:\$11,250$ $=\$15,000\left(.75\right)\:=\:\$11,250$ . At the end of the third year $=\$11,250\left(.75\right)\:=\:\$8,437.50$ $=\$11,250\left(.75\right)\:=\:\$8,437.50$ . At the end of the fourth year $=\$8,437.50\left(.75\right)\:=\:\$6,328.13$ $=\$8,437.50\left(.75\right)\:=\:\$6,328.13$ . At the end of the fifth year $=\$6,328.13\left(.75\right)\:=\:\$4,746.09$ $=\$6,328.13\left(.75\right)\:=\:\$4,746.09$ . Notice that the value is always multiplied by the same rate (.75), so we could model this situation with the function: value (v) of the car $=\$20,000\left(.75\right)^t$ $=\$20,000\left(.75\right)^t$

$V = 20, 000 {(. 75)}^{5}$ $V=20,000\left(.75\right)^5$

$V=\$4,746.09$ $V=\$4,746.09$