17. ( )
4
3 -1 2 1
3
Solve each exponential equation.
1
(a) 2 32 (b)x x x
x
e e
e
− −
= = ×
3 -1 5
2 2x
=
3 1 5x − =
2x =
4
2 1 7
3
x
x x
x
e
e e
e
−
− −
= =
2 1 7x x− = −
1
9
x =
18.
19.
20.
21.
22. A credit union pays interest of 4% per annum compounded quarterly on a
certain savings plan. If $2000 is deposited in such a plan and the interest is
left to accumulate, how much is in the account after 1 year?
( )
1
Interest after first quarter: 2000 0.04 $20
4
I
= = ÷
New Principal =2000 20 $2020+ =
( )
1
Interest after second quarter: 2020 0.04 $20.20
4
I
= = ÷
New Principal =2020 20.20 $2040.20+ =
( )
1
Interest after third quarter: 2040.20 0.04 $20.40
4
I
= = ÷
New Principal =2040.20 20.40 $2060.60+ =
( )
1
Interest after fourth quarter: 2060.60 0.04 $20.61
4
I
= = ÷
New Principal =2060.60 20.61 $2081.21+ =
23.
24.
25.
26.
27.
28. Find the amount A that results from investing a principal P of
$2000 at an annual rate r of 8% compounded continuously for a
time t of 1 year.
( )( )0.08 1
2000eA = ( )( )0.08 1
$22000 16 5e 6. 7= =
29. Suppose you want to open a money market account. You visit
three banks to determine their money market rates. Bank A offers
you 4% annual interest compounded daily, Bank B offers you 4.1%
compounded monthly, and Bank C offers 3.95% compounded
continuously. Determine which bank is offering the best deal.
Bank A Bank B Bank C
365
0.04
1 1
365
er
= + − ÷
0.0400808er =
4.008%er =
12
0.041
1 1
12
er
= + − ÷
0.0417793er =
4.178%er =
0.395
1er e= −
0.0402905er =
4.029%er =
Bank B is offering the best deal
30.
31. What annual rate of interest compounded quarterly should
you seek if you want to double your investment in 6 years?
( )4 6
2 1
4
r
P P
= + ÷
24
2 1
4
r
= +
( )24
4 2 1r = − 0.1171089=
The annual rate of interest needed to
double the principal in 6 years is 11.71%
24
2 1
4
r
= + ÷
32. (a) How long will it take for an investment to double in value
if it earns 6% compounded continuously?
(b) How long will it take to triple at this rate?
0.06
(a) 2 t
P Pe=
0.06
2 t
e=
0.06
ln 2 ln t
e=
ln 2 0.06t=
ln 2
11.55 years
0.06
t = ≈
0.06
(b) 3 t
P Pe=
0.06
3 t
e=
0.06
ln3 ln t
e=
ln3 0.06t=
ln3
18.31 years
0.06
t = ≈
