NAME:
MATH 151
November 12, 2014
QUIZ 8
• Calculators are NOT allowed!
• Show all your work and indicate your final answer clearly. You will be graded not merely
on the final answer, but also on the work leading up to it.
1. (3 points) Differentiate f (x) = log12 (xex ).
Solution:Rewrite f using the change of basis identity and properties of logarithms:
ln(xex )
1
1
f (x) =
=
(ln(x) + ln(ex )) =
(ln(x) + x).
ln(12)
ln(12)
ln(12)
Differentiating this last equation gives
1
1
0
f (x) =
+1
ln(12) x
2. (3 points) Find the slope of the tangent line of y = x10x at x = 1.
Solution: To differentiate y, we use logarithmic differentiation:
ln y = ln(x10x ) = 10x ln(x).
Differentiating the above equality gives
1 0
1
y = 10 ln x + · x = 10(ln(x) + 1)
y
x
so
y 0 = y · 10(ln(x) + 1) = x10x 10(ln(x) + 1).
The slope of the tangent line of y at 1 is
y 0 (1) = 110 · 10 · (ln(1) + 1) = 10 .
(recall ln(1) = 0)
NAME:
MATH 151
November 12, 2014
3. (3 points) Bismuth-210 has a half-life of 5 days. A sample originally has a mass of 800 mg.
Find a formula for the mass remaining after t days.
Solution: The equation for exponential decay is P (t) = P0 ekt where P0 is the initial mass.
So P (t) = 800ekt . What remains is solving for k. Since Bismuth-210 has a half-life of 5
days,
1
1
1
1
5k
5k
400 = 800e =⇒ = e =⇒ ln
= 5k =⇒ k = ln
.
2
2
5
2
Thus
t
1 5
ln
1
t
P (t) = 800e 5 ln( 2 ) = 800e 2
so
5t
1
P (t) = 800
2