MATH 130: CALCULUS FOR THE LIFE SCIENCES I Quiz 1

MATH 130: CALCULUS FOR THE LIFE SCIENCES I Quiz 1

1) Solve the following equation: - = (2)+3

(1 point)

2) Solve the following equation:

3 2-4

=

1

-4

81

(1 point)

3) Given that log 2 = and log 3 = , find log 24 in terms of and . (2 points)

4) Using properties of logarithms, solve log( + 2) + log( + 5) = 1 (2 points)

5) When an antibiotic is introduces into a culture of 50,000 bacteria, the number of bacteria decreases exponentially. After 9 hours, there are only 20,000 bacteria.

a) Write an exponential equation to express the growth function y in terms of

time t in hours.

(2 points)

b) In how many hours will half of the bacteria remain?

(2 points)

In the fifth exercise, you may use the following values of the natural logarithms: ln 0.5 = -0.69 ln 0.4 = -0.92

SOLUTIONS OF QUIZ 1

1.

- = (2)+3 - = 2(+3) - = 2+6 - = 2 + 6 -3 = 6 = -2

2.

3 2-4

=

811 -4

32-4

=

314

-4

3 2-4

=

(3-4 ) -4

3 2-4

=

3-4( -4)

32-4 = 3-4+16 2 - 4 = -4 + 16 2 = 16 = 4

3.

log 24 = log (233) = log 23 + log 3 = 3 log 2 + log 3 = 3 +

4. log( + 2) + log( + 5) = 1 log( + 2)( + 5) = log 10 log(2 + 7 + 10) = log 10

2 + 7 + 10 = 10 2 + 7 = 0 ( + 7) = 0 = 0 = -7

5.

a)

20,000

=

50,0009

9

=

20,000 50,000

=

2 5

9

=

ln 2

5

= ln 0.4

=

-0.92

=

ln 0.4 9

=

-

0.92 9

=

-0.1

Thus, = 50,000(90.4) = 50,000-0.1

b) Suppose we now have N bacteria. We want to calculate the time when we will have N/2 bacteria.

2

=

ln 90.4

ln 90.4

=

1 2

ln 0.4 (

9

)

=

1 ln 2

=

ln 0.5

=

9 ln 0.5 ln 0.4

=

-9 0.69 -0.92

=

6.75

2

=

ln 90.4

ln 90.4

=

1 2

ln 0.4 (

9

)

=

1 ln 2

=

ln 0.5

=

9 ln 0.5 ln 0.4

=

6.81

2

=

-0.1

-0.1

=

1 2

-0.1

=

ln

1 2

=

ln

0.5

=

-0.69

=

-0.69 -0.1

=

6.9

In about 7 hours, the population of the bacteria will be half.

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