MATH 32 SPRING 2013
FINAL EXAM SOLUTIONS
(1) (10 points) Suppose you put $100 in a bank account which gives 6% interest compounded
monthly. If you don’t add or remove money from the account, after how many years will
you have $500 in the account?
You may express your answer in terms of a logarithm - no need to give a decimal approximation.
Recall that the formula for interest compounded n times a year is
r nt
.
A=P 1+
n
Solution: We have P = 100, r = .06, n = 12, and A = 500, and we’d like to find t.
.06 12t
500 = 100 1 +
12
12t
.06
5= 1+
12
12t
ln(5) = ln (1 + .005)
ln(5) = 12t ln(1.005)
t=
ln(5)
12 ln(1.005)
Above I took the natural log, but any other logarithm would have done just as well.
Many students chose to take log1.005 , obtaining the answer
t=
log1.005
.
12
(2) (10 points) Simplify log4 (164 ) + log4 (23 ).
Solution: 4 log4 (16) + 3 log4 (2) = 4(2) + 3( 12 ) = 8 +
3
2
=
19
2 .
(3) (a) (5 points) Find the area of the ellipse given by
(x − 2)2
+ (y − 1)2 = 1.
3
(b) (5 points) Find the circumference of the circle given by
(x − 2)2 + (y − 1)2 = 16.
(c) (5 points) Find the area of the triangle in the plane with vertices (2, −2), (4, −2), and
(1, 2).
Solution:
1
2
MATH 32 SPRING 2013 FINAL EXAM SOLUTIONS
(a) Writing the equation as
(x − 2)2 (y − 1)2
= 1,
√ 2 +
12
3
we see the ellipse has vertical radius 1 and horizontal radius
√
3π.
(b) Writing the equation as
√
√
3, so its area is π(1)( 3) =
(x − 2)2 + (y − 1)2 = 42 ,
we see that the circle has radius 4, so its circumference is 2π(4) = 8π.
(c) Here is a picture of the triangle:
We can see that it has height 4 and base 2, so its area is 21 (2)(4) = 4.
(4) (10 points) Find all values of x satisfying
ln(x) + ln(2x) = 2.
Solution:
ln(x) + ln(2x) = 2
ln(2x2 ) = 2
2x2 = e2
x2 =
e2
2r
x=±
e2
e
= ±√
2
2
But x cannot be negative, since the domain of ln is (0, ∞).
So there is only one solution:
e
x= √ .
2
Alternative solution:
MATH 32 SPRING 2013
FINAL EXAM SOLUTIONS
3
ln(x) + ln(2x) = 2
ln(x) + ln(2) + ln(x) = 2
2 ln(x) = 2 − ln(2)
1
ln(x) = 1 − ln(2)
2
x=e
1+ln( √1 )
2
1
e
x = e( √ ) = √
2
2
(5) Recall that by area(x2 , 1, 2), I mean the area below the graph y = x2 and above the x-axis,
from x = 1 to x = 2.
(a) (10 points) Estimate the value of area(x2 , 1, 2) by computing the area of two rectangles,
each of width 21 , drawn below the graph. Draw a picture to make it clear what area
you are computing.
(b) (5 points) Is the actual value of area(x2 , 1, 2) larger or smaller than your answer in
part (a)?
Solution:
(a) See picture. The first rectangle has area ( 21 )(1) = 12 .
The second rectangle has area ( 21 )( 94 ) = 98 .
Total estimate: 12 + 98 = 48 + 89 = 13
8 .
(b) The actual area is larger.