1. ## Exponential Functions

A bank pays 9 percent interest compounded annually. There are 1000 dollars in the account in January 1933. You may use a calculator.
a) How much money will be in the account in January 1934? dollars
b) How much money will be in the account in January 1935? dollars
c)How much money will be in the account in January 1936? dollars
d) How much money will be in the account in January 1943? dollars
e) How much money will be in the account x years after 1933? dollars f) In what year will there first be one million dollars in the account? [Hint: use your answer from (e) and logs] in the year

2. Originally Posted by lilikoipssn
A bank pays 9 percent interest compounded annually. There are 1000 dollars in the account in January 1933. You may use a calculator.
a) How much money will be in the account in January 1934? dollars
b) How much money will be in the account in January 1935? dollars
c)How much money will be in the account in January 1936? dollars
d) How much money will be in the account in January 1943? dollars
e) How much money will be in the account x years after 1933? dollars f) In what year will there first be one million dollars in the account? [Hint: use your answer from (e) and logs] in the year
Using the Compound interest formula, we have:

$A=P{\left(1+\frac{r}{n}\right)}^{nt}$

Where A is the future amount, P is the principle (initial) amount, r is the interest rate, n is the number of times compounded in a year, and t is the number of years.

a) How much money will be in the account in January 1934? dollars
1934 is 1 year after 1933 (so t=1). Plugging this into the formula we get:

$A=(1000){\left(1+\frac{.09}{1}\right)}^{(1)(1)}=10 90$

b) How much money will be in the account in January 1935? dollars
$A=(1000){\left(1+\frac{.09}{1}\right)}^{(1)(2)}=11 88.10$

c)How much money will be in the account in January 1936? dollars
$A=(1000){\left(1+\frac{.09}{1}\right)}^{(1)(3)}=12 95.03$

d) How much money will be in the account in January 1943? dollars
$A=(1000){\left(1+\frac{.09}{1}\right)}^{(1)(10)}=2 367.36$

e) How much money will be in the account x years after 1933? dollars
$A=1000{\left(1.09\right)}^{x}$

f) In what year will there first be one million dollars in the account? [Hint: use your answer from (e) and logs] in the year
$1000000=1000(1.09)^{t}$
$\Rightarrow 1000=(1.09)^t$
$\Rightarrow ln1000=tln1.09$
$\Rightarrow t=\frac{ln1000}{ln1.09}\approx 80.16 years$

Hoped this helped you out!!