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

Write

      2 log_q (5x)  - log_q (3x - 2) as a single logarithm

Solution

First use the power rule to get

       log_q (5x)²  - log_q (3x - 2)

      =  log_q (25x²) - log_q (3x - 2)

Next use the quotient property to get

                   25 x² 
        log_q                  
                  3x - 2

Problem 5

Expand the logarithm 

                  25 x¹⁰ 
        log₅                    
                    y¹²

as much as possible and simplify.

Solution

First use the quotient property of logs to get

        log₅(25x¹⁰)  -  log₅(y¹²)

Now use the product property on the first term to get

        log₅(25)  +  log₅(x¹⁰)  -  log₅(y¹²)

Now use the power property and the fact that 25 = 5² so that the first term is just 2.

        2  +  10 log₅(x)  -  12 log₅(y)

Problem 6

A. Find the equation of the conic shown below

Solution

Since the vertices are at (-4,0), (4,0) and (0,-2), (0,2), the conic is an ellipse with a = 4 and b = 2.  The equation is

             x²            y²    
                     +                =  1
            16            4

B.    Sketch the graph of the conic with the equation below:

             9y²   -   x²   =  36

Solution

First divide all three terms by 100 to get

             y²            x²    
                     -                =  1
            4             36

Now notice that this is a hyperbola with a = 6 and b = 2.  Now sketch the hyperbola.  The vertices of the hyperbola are at

(0,-2) and (0,2).  The helper points to complete the rectangle are at (-6,0) and (6,0).  If you sketch the rectangle and use the diagonals as the asymptotes, you will get the graph shown below.

Problem 7

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 8 

A.  Find the indicated unknown

        8, 13, 18, 23, ..., 88            n  =  ?

Solution

This is an arithmetic sequence since 

        13 - 8  = 18 - 3  =  23 - 18  =  5

The common difference is d  =  5.

The first term is 8, so the general formula is

        a_n  =  8 + 5(n - 1)

The last term is 88, so we set

        88  =  8 + 5(n - 1)    

        80  =  5(n - 1)    Subtract 8

        16  =  n - 1        divide by 5

        n  =  17        add 1

B.  If the second term of a geometric sequence is -2/3 and the third term is 1/3, find a₁ and r.

Solution

Since this is a geometric sequence, we can find r by dividing a term by its previous term.  That is

                 1/3             1         3           1
r  =                =            x          = -                                
           -2/3             3        -2           2
                   

Now we can find a₁ by using the formula

        a₂ = a₁ (r)¹

or

        -2/3  =  a₁(-1/2)

so that

                 -2/3             -2         -2          4
a₁  =                =            x          =                                   
             -1/2             3          1           3
 

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

Problem 11

Lake Tahoe Community College had 4200 students in 2002, and has had a enrollment decline of 150 students each year since 2002. 

A.  Find the enrollment for each of five years, beginning with 2002.

Solution

To find the enrollments, just subtract 150 from the previous number.  The enrollments are shown below:

B.  Find the general term a_n of this sequence.

This is just an arithmetic sequence with first term 4200 and common difference -150.  The general term is

        a_n = a₁ + d(n - 1)  =  4200 + (-150)(n-1)

        a_n  =  4200 - 150(n - 1)