1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
October 31
1. If loga 2 = 0.37, log a 3 = 0.56 and loga 5 = 0.83, then evaluate loga
2
6a2
25
loga 6a
= loga 6a2 − loga 25 = loga 6 + loga a2 − loga 25 = loga 2 ∗ 3 + 2 loga a − loga 52
25
= loga 2 + loga 3 + 2 − 2 loga 5 = 0.37 + 0.56 + 2 − 2(0.83) = 1.27
2. Solve for x.
(a) logx (8 − 2x) = 2
First rewrite the problem as an exponential.
x2 = 8 − 2x
x2 + 2x − 8 = 0
(x + 4)(x − 2) = 0
x = −4 or x = 2.
Since the base of a logarithm can not be zero or negative, then the only solution is x = 2
(b) log6 (x − 2) + log6 (x + 7) = 2
First combine the logs on the left side of the equation and then rewrite the problem as an
exponential.
log6 ((x − 2)(x + 7)) = 2
62 = (x − 2)(x + 7)
36 = x2 + 5x − 14
x2 + 5x − 50 = 0
(x + 10)(x − 5) = 0
x = −10 or x = 5.
Since you can not take a logarithm of a negative number the only solution is x = 5.
2
(c) 7x = 20
Take the natural log of both sides. (Note: you can actually use any logarithm.)
2
ln 7x = ln 20
x2 ln 7 = ln 20
x2 = lnln20
7
Answer x = ±
q
ln 20
ln 7
≈ ±1.53950185
3. Take the derivative of the function and completely simplify.
y=
(x + 1)7 (2x + 1)6
x4
My advice is to take a natural log of both sides and simplify with the logarithm rules.
ln y
ln y
ln y
ln y
7
6
= ln (x+1) x(2x+1)
4
= ln (x + 1)7 (2x + 1)6 − ln x4
= ln(x + 1)7 + ln(2x + 1)6 − ln x4
= 7 ln(x + 1) + 6 ln(2x + 1) − 4 ln x
now take a derivative and then simplify.
y0
1
2
1
=7∗
+6∗
−4∗
y
x+1
2x + 1
x
2
151 WebCalc Fall 2002-copyright Joe Kahlig
y0 = y
7
12
4
+
−
x + 1 2x + 1 x
(x + 1)7 (2x + 1)6 7x(2x + 1) + 12x(x + 1) − 4(x + 1)(2x + 1)
y =
x4
x(x + 1)(2x + 1)
0
"
(x + 1)7 (2x + 1)6 14x2 + 7x + 12x2 + 12x − 8x2 − 12x − 4
y =
x4
x(x + 1)(2x + 1)
#
0
y0 =
(x + 1)6 (2x + 1)5 2
18x
+
7x
−
4
x5
4. Take the derivative of the following.
(a) y = log4 x4 + 5 sin(x)
4x3 + 5 cos(x)
y0 = 4
(x + 5 sin(x)) ln(4)
(b) y = (2x + 1)sin x
First take the natural log
of both sides.
ln y = ln (2x + 1)sin(x)
ln y = sin(x) ln(2x + 1)
y0
2
= cos(x) ln(2x + 1) + sin(x) ∗
y
2x + 1
2 sin x
0
sin x
y = (2x + 1)
cos(x) ln(2x + 1) +
2x + 1
5. A bacteria culture starts with 5000 critter and triples every three hours.
(a) Find a formula, assuming exponential growth, that gives the number of critters in the
culture after t hours.
The points that we use are (0, 5000) and (3, 15000) and the formula is y = Ao ekt with
Ao = 5000
15000 = 5000e3k
3 = e3k
ln(3) = 3k
k = ln(3)
3 ≈ 0.3662040962
Answer y = 5000e0.3662040962t
(b) How many critters are in the culture after 60 minutes?
plug in a 1 for t and you get y = 7211.24785. Answer: approximately 7221
(c) How long until there are 45,000 critters?
45000 = 5000ekt
9 = ekt
ln(9) = kt
t = ln(9)
k = 6 hours
(d) How long until the number of critters doubles?
10000 = 5000ekt
2 = ekt
ln(2) = kt
t = ln(2)
k ≈ 1.892789 hours