Practice Exam 2
Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your work.
Problem 1
Use your calculator to find log329 rounded to three decimal places.
Solution
We use the change of base formula.
log 29 1.4624
log3 x
=
=
= 3.065
log
3
0.4771
Problem 2
Solve for x: ln(x + 1) + ln(x + 2) = 3
Solution
First use the sum to product formula to write this as a single logarithm.
ln[(x + 1)(x + 2)] = 3
Now write this as an exponent
(x + 1)(x + 2) = e3
Multiply the left hand side out to get
x2 + 3x + 2 = e3
or
x2 + 3x + 2 - e3
This is a quadratic with
a = 1 b = 3 c = 2 - e3
We can use the quadratic formula to get
Problem 3
The number of grams N of C14 left in a fossil after t years is
N = 0.03e-.000121t
A fossil has been analyzed and found to contain 0.0002 grams. How old is the fossil?
Solution
We have
0.0002 = 0.03e-.000121t
0.0066667 = e-.000121t
Now turn this into a natural logarithm equation
-0.000121t = ln(0.0066667) = -5.01
Divide by -0.000121 to get
t = 41,410 years old
Problem 4
Solve the following equations
A.
Solution
Square both sides to get
FOIL the left hand side to get
or
Next subtract x + 23 from both sides to get
Square both sides to get
100(x - 2) = x2 + 28x + 196
100x - 200 = x2 + 28x + 196
x2 - 72x + 396 = 0
(x - 6)(x - 66) = 0 If you have trouble factoring, you can use the quadratic formula.
Hence
x = 6 or x = 66
Plugging into the original equation verifies that these are indeed the two solutions.
B. (3x - 2)2 +9(3x - 2) = -20
Let
u = 3x - 2
and add 20 to both sides to get
u2 + 9u + 20 = 0
or
(u + 4)(u + 5) = 0
u = -4 or u = -5
so that
3x - 2 = -4 or 3x -2 = -5
3x = -2 or 3x = -3
x = -2/3 or x = -1
Problem 5
Solve the following nonlinear system of equations
2x2 - 3y2 = 29
x2 + y2
= 17
Solution
We can multiply the second equation by 2 to get
2x2 - 3y2 = 29
2x2 + 2y2
= 34
Now subtract the second equation from the first to get
-5y2 = -5
y2 = 1
y = -1 or y = 1
Plugging these into the second equation gives
2x2 + 2 = 34
x2 + 1 = 17
x2 = 16
x = -4 or x = 4
This gives the four points
(-4,-1), (-4,1), (4,-1) and (4,1)
Problem 6
Use the graphs below to solve the following nonlinear system
x2 + y4 = 1
x + y = 3
Solution
Since the graphs do not intersect, the system has no solution.
Problem 7
Solve the inequalities
A. x2 + 3x - 4 < 0
Solution
First factor
(x - 1)(x + 4) < 0
The two key points are
x = -4 and x = 1
Now we use test values:
When x = -5,
(-5 - 1)(-5 + 4) > 0
so the statement is false.
When x = 0,
(0 - 1)(0 + 4) < 0
so the statement is true.
When x = 2,
(2 - 1)(2 + 4) > 0
so the statement is false.
Hence the solution is the interval (-4,1).
B.
x-2
> 2
x-5
First subtract 2 from both sides to get
x
- 2
-
2 > 0
x - 5
x
- 2 2x - 10
-
> 0
x -
5 x - 5
x-2
- 2x + 10
>
0
x
- 5
-x
+ 8
> 0
x-5
The two key points are
x = 8 and x = 5
Now test points in the three regions.
When x = 0,
0
+ 8
< 0
0 - 5
and the statement is false.
When x = 6,
-6
+ 8
> 0
6 - 5
and the statement is true.
When x = 9,
9
+ 8
< 0
-9 - 5
and the statement is false.
Hence the solution is
(5,8]
Notice that 8 is included because of the ">" sign, while 5 is not since the function is undefined at x = 5.
Problem 8
The graph of the function
x2 + x + 4
y =
x - 2
is shown below. Use the graph to solve the inequality
x2 + x + 4
< 0
x - 2
Solution
We see that the graph is above the x-axis for values of x less than 2. Hence the solution set is
(-
,
2)
Problem 9
Sketch the graph of the circle with equation
x2 + y2 + 4x - 6y = 3
We first complete the squares
x2 + 4x + 4 + y2 - 6y + 9 = 3 + 4 + 9
(x + 2)2 + (y - 3)2 = 16
Hence the center is at (-2,3) and the radius is 4. The graph is shown below.
Problem 10
Find the equation of the parabola that opens downward, has vertex (1,3) and passes through the point (2,1).
Solution
The standard equation is
y = a(x - h)2 + k
Since the vertex is at (1,3) we have
y = a(x - 1)2 + 3
Since it passes through the point (2,1), we have
1 = a(2 - 1)2 + 3 = a + 3
hence
a = -2
The parabola has equation
y = -2(x - 1)2 + 3 = -2(x2 - 2x + 1) + 3
y = -2x2 + 4x + 1