MATH 147
Lab Key #2
2/2/2016
(1) Simplify the following expressions completely:
(a) 27−2 log3 x
(b) log2 (16− x )
(c) log x2 − 1 − log ( x + 1)
2
(d) e− ln( x +1)
Solution: (a)
−2 log3 x
−6
1
27−2 log3 x = 33
= 3−6 log3 x = 3log3 ( x ) = x −6 = 6 .
x
(b)
log2 16
−x
−x
= log2 2−4x = −4x.
= log2 24
(c)
log x2 − 1 − log ( x + 1) = log [( x + 1) ( x − 1)] − log ( x + 1)
= log ( x + 1) + log ( x − 1) − log ( x + 1) = log ( x − 1) .
(d)
e− ln( x
2 +1
h
) = eln ( x2 +1)
−1
i
−1
= x2 + 1
=
1
.
x2 + 1
1
(2) A common inhabitant of human intestines is the bacterium Escherichia coli.
A cell of this bacterium in a nutrient broth medium divides into two cells
every 20 minutes.
(a) If the initial population of a culture is 100 cells, find an expression for the
number of cells y (t) after t hours. [Hint: The doubling time is 20 minutes =
1/3 hour.]
(b) When will the population reach 1000 cells? Express your answer in terms
of the natural logarithm and include the appropriate unit for time.
Solution: (a) We will look for an expression of the form y (t) = y0 ekt , where
k is a constant and y0 is the initial population. Therefore,
P = 100. Since the
population doubles every third of an hour, that mean y
1
3
= 200. Hence,
200 = 100ek/3 .
∴ 2 = ek/3 .
k
∴ ln 2 = .
3
Hence, k = 3 ln 2 = ln 8 and so y (t) = 100 · e3t ln 2 = 100 · et ln 8 .
(b) Now, we are looking for when y (t) = 10000. Therefore, we want to
solve the following equation for t :
100 · e3t ln 2 = 10000.
∴ e3t ln 2 = 100.
∴ 3t ln 2 = ln (100) .
∴t=
ln (100)
ln (100)
hours =
hours.
3 ln (2)
ln (8)
(3) After 3 days a sample of radon-222 decayed to 58% its original amount.
(a) What is the half-life of radon-222?
(b) How long will it take a sample to decay to 10% of its original amount.
Solution: (a) First we attempt to find a formula of the form y (t) = y0 ekt for
the amount of radon-222 left over after t days, where y0 is the original amount.
Then y (3) is 58% of P. Therefore,
.58P = y (3) = y0 e3k .
∴ .58 = e3k .
∴ ln (0.58) = 3k.
2
Therefore, k = 13 ln (0.58) . Hence, y (t) = y0 e(t/3) ln(0.58) . Now, let t represent
the half-life of radon-222. Then y (t) = 0.5y0 .
∴ 0.5y0 = y (t0 ) = y0 e(t/3) ln(0.58) .
∴ 0.5 = e(t/3) ln(0.58) .
t
∴ ln (0.5) = 0 ln (0.58) .
3
ln (0.5)
∴ t0 = 3
days.
ln (0.58)
(b) Now we are interested in when y (t) = 0.1y0 . Therefore,
0.1y0 = y0 e(t/3) ln(0.58) .
∴ 0.1 = e(t/3) ln(0.58) .
t
∴ ln (0.1) = ln (0.58) .
3
ln (0.1)
days.
∴t=3
ln (0.58)
(4) Use the double log plot provided to find the functional relationship between x and y.
Solution: Changing to logarithm coordinates (log x, log y) , we have a line
passing through (1, 8) and (2, 2) . The slope of this line is given by
m=
2−8
= −6.
2−1
Hence,
log y − 2 = −6 (log x − 2) .
∴ log y − 2 = −6 log x + 12.
∴ log y = −6 log x + 14.
Now, to solve for y, we proceed as follows:
10log y = 10−6 log x+14 .
−6
∴ y = 10log( x ) · 1014 .
∴ y = 1014 x −6 .
∴y=
1014
.
x6
3