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MATH 154 PRACTICE MIDTERM II

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Do all your work and give all your answers on you own sheet of paper.  Show your work.

Problem 1 

Solve the following:

A.     (10 Points)   log₆(x - 5) + log₆(x)  =  2      

Solution

Use the sum to product rule of logarithms

        log₆[(x)(x - 5)]  =  2    

        (x)(x - 5)  =  6²    Remember logs are exponents

        x² - 5x  =  36        Multiplying out

        x² - 5x - 36  =  0      Subtracting 36 from both sides

        (x - 9)(x + 4)  =  0

        x  =  9    or     x  =  4

Notice that 4 is not a solution since it is not in the domain of log₆(x - 5).  9 is a solution, as you can verify with your calculator.

B.  (10 Points)  2^2x-1  =  5^3x-2  

Solution

Take ln of both sides

        ln 2^2x-1  =  ln 5^3x-2  

        (2x + 1) ln 2  =  (3x - 2) ln 5    Using the power to product rule of logs

        2x ln 2  +  ln2    =    3x ln 5  -  2 ln 5    Multiplying out

        2x ln 2  -  3x ln 5    =    -ln 2  -  2 ln 5    Separating x terms to the left and others to the right

        x(2 ln 2  -  3 ln 5)    =    -ln 2  -  2 ln 5    Factoring out the x

                  -ln 2  -  2 ln 5
        x  =                                  =  1.1365    Dividing by 2 ln 2  -  3 ln 5  
                   2 ln 2  -  3 ln 5

C.  (10 Points)   Use a calculator to evaluate  log₇ 51

Solution

First use the change of base formula

                            log 51  
        log₇ 51  =                      
                            log 7  

Now put this in your calculator to get

        2.02055867514

Problem 2 

The clarity C (in feet) of Lake Tahoe t  years since 1990 can be modeled by the equation

        C(t)  =  45e^-t/25

A.     (9 Points)   How clear was the lake in 1995.

Solution

The year 1995 corresponds to t  =  5.  Plug this in for t to get

        C(5)  =  45e^-5/25  =  36.84

The clarity of the lake was 36.84 feet in 1995

B.     (10 Points)   When will the clarity of the lake only be 5 feet?

Solution

Now plug in 5 for C to get

        5  =  45e^-t/25

        1/9  =  e^-t/25        Dividing both sides by 45

        ln(1/9)  =  ln(e^-t/25)      Taking the ln of both sides

        ln(1/9)  =  -t/25        Cancelling the ln and the e

        -25 ln(1/9)  =  t        Multiplying both sides by -25

        t  =  54.93

Now turn this back into a date by adding it to 1990

        1990 + 54.93  =  2044.93

We can say that towards the end of the year 2044 the clarity of the lake will be only 5 feet.

Problem 3   

One muffin, two pies, and three cakes cost $23.  One Muffin, three pies , and two cakes cost $21.  One muffin, four pies, and five cakes cost $39.  Find the cost of each.

Solution

Let

        x  =  the cost per muffin

        y  =  the cost per pie

        z  =  the cost per cake

We get three equations:

        x + 2y + 3z  =  23

        x + 3y + 2z  =  21

        x + 4y + 5z  =  39

Subtracting the second equation from the first gives

        -y + z  =  2

Subtracting the second equation from the third gives

        y + 3z  =  18

Now add the two equations above to get

        4z  =  20

        z  =  5    dividing by 4

Plugging z  =  5 back into the equation above to get

        y + 3(5)  =  18

        y  =  3        subtracting by 15

Now plug in y  =  3 and z  =  5 into the first original equation to get

        x + 2(3) + 3(5)  =  23

        x + 21  =  23            6 + 15  =  23

        x  =  2                       Subtracting 2 from both sides

Now plug in (2,3,5) into the second and third equations to check.

        2 + 3(3) + 2(5)  =  21

        2 + 4(3) + 5(5)  =  39

We can conclude that the muffins cost $2, the pies cost $3, and the cake costs $5.

Problem 6  Expand each logarithm.  Simplify where possible.  Assume all variables are such that all expressions are defined.

A.     (10 Points)                 x⁵y²
                        log₃                
                                   81

Solution

Use the quotient to difference property of logarithms first:

        log₃(x⁵y²) - log₃(81)

        =   log₃(x⁵) + log₃(y²) - log₃(3⁴)        We used the product to sum property.

        =   5 log₃(x) + 2 log₃(y) - 4        We used the product to sum property and cancelled the log₃ and 3

B.     (10 Points)         

Solution

Use the product to sum property of logarithms first and change the root into an exponent. 

Warning:  Do not cancel the root with the squares, since the "+" is in the way.

        log₅(x) + log₅[(y² + 25)^1/2]

        =  log₅(x) + 1/2 log₅(y² + 25)

Problem 7 Answer the following True or False.  If True, explain your reasoning, if False, explain your reasoning or show a counter-example.

A.     (7 Points)   The domain of the function  y  =  log_b x  is (0,  )  for all b > 0.

Solution

True,  the domain of the log is the range of the exponential which takes on all positive numbers

B.     (7 Points)   If a linear system of three equations with three unknowns has (1,2,3) as a solution then (4,5,6) cannot also be a solution

Solution

False,  it is possible for three planes to intersect in a line, which could contain both points.

C.     (7 Points)  If the vertex of a parabola has y-coordinate 3 and the directrix is y = 2, then the parabola has no x-intercepts.

Solution

True,  all of the parabola lies on the same side of the directrix as the vertex, hence the smallest y could get is 2.  An x-intercept has y coordinate 0.

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